Virtual University of Pakistan

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1 Calculus II MTH31 Virtual University of Pakistan Knowledge beyond the boundaries 1

2 Table of Contents Lecture No. 1 Introduction....3 Lecture No. Values of function..7 Lecture No. 3 Elements of three dimensional geometry...11 Lecture No. 4 Polar co-ordinates...17 Lecture No. 5 Limits of multivariable function....4 Lecture No. 6 Geometry of continuous functions Lecture No. 7 Geometric meaning of partial deri.vative Lecture No. 8 Euler theorm Lecture No.9 Examples Lecture No. 1 Introduction to vectors Lecture No. 11 The triple scalar product or Box product...6 Lecture No. 1 Tangent Planes to the surfaces 69 Lecture No. 13 Noarmal Lines Lecture No. 14 Extrema of function of two variables Lecture No. 15 Examples Lecture No. 16 Extreme Valued Theorm Lecture No. 17 Example 98 Lecture No. 18 Revision Of Integration..15 Lecture No. 19 Use of integrals Lecture No. Double integral for non-rectangular region..113 Lecture No. 1 Examples Lecture No. Examples Lecture No. 3 Polar co-ordinate systems...14 Lecture No. 4 Sketching Lecture No. 5 Double Integrals in Polar co-ordinates Lecture No. 6 Examples Lecture No. 7 Vector Valued Functions.. 14 Lecture No. 8 Limits Of Vector Valued Functions..144 Lecture No. 9 Change Of Parameter..15 Lecture No. 3 Exact Differential..155 Lecture No. 31 Line Intagral Lecture No. 3 Examples Lecture No. 33 Examples Lecture No. 34 Examples Lecture No. 35 Definite Lecture No. 36 Scalar Field Lecture No. 37 Examples Lecture No. 38 Vector Filed Lecture No. 39 Periodic Functions..197 Lecture No. 4 Fourier Series....1 Lecture No. 41 Examples..6 Lecture No. 4 Examples..1 Lecture No. 43 Functions with periods other than Lecture No. 44 Laplace Transforms.. Lecture No. 45 Theorems...7

3 1-Introduction Lecture No-1 Introduction Calculus is the mathematical tool used to analyze changes in physical quantities. Calculus is also Mathematics of Motion and Change. Where there is motion or growth, where variable forces are at work producing acceleration, Calculus is right mathematics to apply. Differential Calculus Deals with the Problem of Finding (1)Rate of change. ()Slope of curve. Velocities and acceleration of moving bodies. Firing angles that give cannons their maximum range. The times when planets would be closest together or farthest apart. Integral Calculus Deals with the Problem of determining a Function from information about its rates of Change. Integral Calculus Enables Us (1) To calculate lengths of curves. () To find areas of irregular regions in plane. (3) To find the volumes and masses of arbitrary solids (4) To calculate the future location of a body from its present position and knowledge of the forces acting on it. Reference Axis System Before giving the concept of Reference Axis System we recall you the concept of real line and locate some points on the real line as shown in the figure below, also remember that the real number system consist of both Rational and Irrational numbers that is we can write set of real numbers as union of rational and irrational numbers. Here in the above figure we have locate some of the rational as well as irrational numbers and also note that there are infinite real numbers between every two real numbers. Now if you are working in two dimensions then you know that we take the two mutually perpendicular lines and call the horizontal line as x-axis and vertical line as y-axis and where these lines cut we take that point as origin. Now any point on the x-axis will be denoted by an order pair whose first element which is also known as abscissa is a real number and other element of the order pair which is also known as ordinate will has values. Similarly any point on the y-axis can be representing by an order pair. Some points are shown in the figure below. Also note that these lines divide the plane into four regions, First,Second,Third and Fourth quadrants respectively. We take the positive real numbers at the right side of the origin and negative to the left side, in the case of x-axis. Similarly for y-axis and also shown in the figure. 3

4 1-Introduction Location of a point Now we will illustrate how to locate the point in the plane using x and y axis. Draw two perpendicular lines from the point whose position is to be determined. These lines will intersect at some point on the x-axis and y-axis and we can find out these points. Now the distance of the point of intersection of x-axis and perpendicular line from the origin is the X-C ordinate of the point P and similarly the distance from the origin to the point of intersection of y-axis and perpendicular line is the Y-coordinate of the point P as shown in the figure below. In space we have three mutually perpendicular lines as reference axis namely x,y and z axis. Now you can see from the figure below that the planes x=,y= and z= divide the space into eight octants. Also note that in this case we have (,,) as origin and any point in the space will have three coordinates. 4

5 1-Introduction Sign of co-ordinates in different octants First of all note that the equation x= represents a plane in the 3d space and in this plane every point has its x-coordinate as, also that plane passes through the origin as shown in the figure above. Similarly y= and z= are also define a plane in 3d space and have properties similar to that of x=.such that these planes also pass through the origin and any point in the plane y= will have y-coordinate as and any point in the plane z= has z-coordinate as. Also remember that when two planes intersect we get the equation of a line and when two lines intersect then we get a plane containing these two lines. Now note that by the intersection of the planes x= and z= we get the line which is our y-axis. Also by the intersection of x= and y= we get the line which is z-axis, similarly you can easily see that by the intersection of z= and y= we get line which is x-axis. Now these three planes divide the 3d space into eight octants depending on the positive and negative direction of axis. The octant in which every coordinate of any point has positive sign is known as first octant formed by the positive x, y and z axis. Similarly in second octant every points has x-coordinate as negative and other two coordinates as positive correspond to negative x- axis and positive y and z axis. Now one octant is that in which every point has x and y coordinate negative and z- coordinate positive, which is known as the third octant. Similarly we have eight octants depending on the sign of coordinates of a point. These are summarized below. First octant (+, +, +) Formed by positive sides of the three axis. Second octant (-, +, + ) Formed by ve x-axis and positive y and z-axis. Third octant ( -, -, +) Formed by ve x and y axis with positive z-axis. Fourth octant ( +, -, +) Formed by +ve x and z axis and ve y-axis. Fifth octant (+, +, -) Formed by +ve x and y axis with -ve z-axis. Sixth octant ( -, +, -) Formed by ve x and z axis with positive y-axis. Seventh octant ( -, -, -) Formed by ve sides of three axis. Eighth octant ( +, -, -) Formed by -ve y and z-axis with +ve x-axis. (Remember that we have two sides of any axis one of positive values and the other is of negative values) Now as we told you that in space we have three mutually perpendicular lines as reference axis. So far you are familiar with the reference axis for d which consist of two perpendicular lines namely x-axis and y-axis. For the reference axis of 3d space we need another perpendicular axis which can be obtained by the cross product of the two vectors, now the direction of that vector can be obtained by Right hand rule. This is illustartaed below with diagram. 5

6 1-Introduction Concept of a Function Historically, the term, function, denotes the dependence of one quantity on other quantity. The quantity x is called the independent variable and the quantity y is called the dependent variable. We write y = f (x) and we read y is a function of x. The equation y = x defines y as a function of x because each value assigned to x determines unique value of y. Examples of function The area of a circle depends on its radius r by the equation A= r so, we say that A is a function of r. The volume of a cube depends on the length of its side x by the equation V= x 3 so, we say that V is a function of x. The velocity V of a ball falling freely in the earth s gravitational field increases with time t until it hits the ground, so we say that V is function of t. In a bacteria culture, the number n of present after one day of growth depends on the number N of bacteria present initially, so we say that N is function of n. Function of Several Variables Many functions depend on more than one independent variable. Examples The area of a rectangle depends on its length l and width w by the equation A = l w, so we say that A is a function of l and w. The volume of a rectangular box depends on the length l, width w and height h by the equation V = l w h so, we say that V is a function of l, w and h. The area of a triangle depends on its base length l and height h by the equation A= ½ l h, so we say that A is a function of l and h. The volume V of a right circular cylinder depends on its radius r and height h by the equation V= rh so, we say that V is a function of r and h. Home Assignments: In the first Lecture we recall some basic terminologies which are essential and prerequisite for this course. You can find the Home Assignments on the last page of Lecture # 1 at LMS. 6

7 -Values of functions Lecture No- Values of functions: Consider the function f(x) = x 1, then f(1) = (1) 1 = 1, f(4) = (4) 1 = 31, f(-) = (-) 1 = 7 f(t-4) = (t-4) 1= t -16t + 31 These are the values of the function at some points. Example Now we will consider a function of two variables, so consider the function f(x,y) =x y+1 then f(,1) =( )1+1=5, f(1,) =(1 )+1=3, f(,) =( )+1=1, f(1,-3) =(1 )(-3)+1=-, f(3a,a) =(3a) a+1=9a 3 +1, f(ab,a-b) =(ab) (a-b)+1=a 3 b -a b 3 +1 These are values of the function at some points. Example: 3 Now consider the function f ( xy, ) = x+ xythen (a) 3 3 f (,4) = + ()(4) = + 8 = + = (b) f (, tt) = t+ ()( t t) = t+ t = t+ t= t (c) f ( xx, ) = x+ ( x)( x) = x+ x = x+ x= x 3 (d) f ( y,4 y) = y + 3 ( y )(4 y) = y + 3 8y = y + y Example: Now again we take another function of three variables f ( xyz,, ) = 1 x y z Then f (,, ) = 1 ( ) ( ) = Example: Consider the function f(x,y,z) =xy z 3 +3 then at certain points we have f(,1,) =()(1) () 3 +3=19, f(,,) =()() () 3 +3=3, f(a,a,a) =(a)(a) (a) 3 +3=a 6 +3 f(t,t,-t) =(t)(t ) (-t) 3 +3=-t 8 +3, f(-3,1,1) =(-3)(1) (1) 3 +3= Example: Consider the function f(x,y,z) =x y z 4 where x(t) =t 3 y(t)= t and z(t)=t (a) f(x(t),y(t),z(t)) =[x(t)] [y(t)] [z(t)] 4 =[ t 3 ] [t ] [t] 4 = t 14 (b) f(x(),y(),z()) =[x()] [y()] [z()] 4 =[ 3 ] [ ] [] 4 = Example: Let us consider the function f(x,y,z) = xyz + x then f(xy,y/x,xz) = (xy)(y/x)(xz) + xy = xy z+xy. Example: Let us consider g(x,y,z) =z Sin(xy), u(x,y,z) =x z 3, v(x,y,z) =Pxyz, xy wxyz (,, ) = Then. z g(u(x,y,z), v(x,y,z), w(x,y,z)) = w(x,y,z) Sin(u(x,y,z) v(x,y,z)) Now by putting the values of these functions from the above equations we get g(u(x,y,z), v(x,y,z), w(x,y,z)) = xy z Sin[(x z 3 )( Pxyz)] = xy z Sin[(Pyx3 z 4 )]. 7

8 -Values of functions Example: Consider the function g(x,y) =y Sin(x y) and u(x,y) =x y 3 v(x,y) = xy Then g(u(x,y), v(x,y)) = v(x,y) Sin([u(x,y) ] v(x,y)) By putting the values of these functions we get g(u(x,y), v(x,y)) = xy Sin([x y 3 ] xy) = xy Sin(x 5 y 7 ). Function of One Variable A function f of one real variable x is a rule that assigns a unique real number f( x ) to each point x in some set D of the real line. Function of two Variables A function f in two real variables x and y, is a rule that assigns unique real number f (x,y) to each point (x,y) in some set D of the xy-plane. Function of three variables: A function f in three real variables x, y and z, is a rule that assigns a unique real number f (x,y,z) to each point (x,y,z) in some set D of three dimensional space. Function of n variables: A function f in n variable real variables x 1,x, x 3,, x n, is a rule that assigns a unique real number w = f(x 1, x, x 3,, x n ) to each point (x 1, x, x 3,, x n ) I n some set D of n dimensional space. Circles and Disks: PARABOLA Parabola y = -x 8

9 -Values of functions General equation of the Parabola opening upward or downward is of the form y = f(x) = ax +bx + c. Opening upward if a >. Opening downward if a <. x co-ordinate of the vertex is given by x = -b/a. So the y co-ordinate of the vertex is y = f(x ) axis of symmetry is x = x. As you can see from the figure below Sketching of the graph of parabola y = ax +bx + c Finding vertex: x co-ordinate of the vertex is given by x = - b/a So, y co-ordinate of the vertex is y = a x +b x + c. Hence vertex is V(x, y ). Example: Sketch the parabola y = - x + 4x Solution: Since a = -1 <, parabola is opening downward. Vertex occurs at x = - b/a = (-4)/(-1) =. Axis of symmetry is the vertical line x =. The y-co-ordinate of the vertex isy = -() + 4() = 4. Hence vertex is V(, 4 ). The zeros of the parabola (i.e. the point where the parabola meets x-axis) are the solutions to -x +4x = so x = and x = 4. Therefore (,)and (4,) lie on the parabola. Also (1,3) and (3,3) lie on the parabola. Graph of y = - x + 4x Example y = x - 4x+3 Solution: Since a = 1 >, parabola is opening upward.vertex occurs at x = - b/a = (4)/ =.Axis of symmetry is the vertical line x =. The y co-ordinate of the vertex is y = () - 4() + 3 = -1.Hence vertex is V(, -1 )The zeros of the parabola (i.e. the point where the parabola meets x-axis) are the solutions to x - 4x + 3 =, so x = 1 and x = 3.Therefore (1,)and (3,) lie on the parabola. Also (,3 ) and (4, 3 ) lie on the parabola. 9

10 -Values of functions Graph of y = x - 4x+3 Ellipse Hyperbola Home Assignments: In this lecture we recall some basic geometrical concepts which are prerequisite for this course and you can find all these concepts in the chapter # 1 of your book Calculus By Howard Anton. 1

11 3-Elements of three dimensional geometry Lecture No-3 Elements of three dimensional geometry Distance formula in three dimension Let Px ( 1, y1, z1) and Qx (, y, z ) be two points such that PQ is not parallel to one of the coordinate axis Then PQ = ( x x ) + ( y y ) + ( z z ) Which is known as Distance fromula between the points P and Q. Example of distance formula Let us consider the points A (3,, 4), B (6, 1, 1), and C (9, 4, 1) Then AB = (6 3 ) + (1 ) + ( 1 4) = 98 = 7 AC = (9 3 ) + (4 ) + (1 4) = 49 = 7 BC = (9 6 ) + (4 1 ) + (1 + 1) = 49 = 7 Mid point of two points If R is the middle point of the line segment PQ, then the co-ordinates of the middle points are x= (x1+x)/, y= (y1+y)/, z= (z1+z)/ Let us consider tow points A(3,,4) and B(6,1,-1) Then the co-ordinates of mid point of AB is [(3+6)/,(+1)/,(4-1)/] = (9/,6,3/) Direction Angles The direction anglesα β, γ of a lineare defined as α = Angle between lineand the positive x-axis β = Angle between line and the positive y-axis γ = Angle between lineand the positive z-axis. By definition, each of these angles lies between. and Direction Ratios Cosines of direction angles are called direction cosines Any multiple of direction cosines are called direction numbers or direction ratios of the line L. Given a point, finding its Direction cosines y-axis 11

12 3-Elements of three dimensional geometry r β P(x,y) y From triangle we can write cos α = x/r cos β = y/r O α x x-axis Direction angles of a Line The angles which a line makes with positive x,y and z-axis are known as Direction Angles. In the above figure the blue line has direction angles as α, and which are the angles which blue line makes with x,y and z-axis respectively. Direction cosines: Now if we take the cosine of the Direction Angles of a line then we get the Direction cosines of that line. So the Direction Cosines of the above line are given by cos α = x OP = x x + y + z cos β = Similarly, cos γ = y OP z OP = = y x + y + z z x +y + z c os α + cos β + cos γ = 1. Direction cosines and direction ratios of a line joining two points For a line joining two points P(x 1, y 1, z 1 ) and Q(x, y, x ) the direction ratios are 1

13 3-Elements of three dimensional geometry x - x 1 y - y 1 z - z1 x - x 1, y - y 1, z - z 1 and the directions cosines are, and PQ PQ PQ. Example For a line joining two points P(1,3,) and Q(7,-,3) the direction ratios are 7-1, - 3, 3 6, -5, 1 and the directions cosines are 6/ 6, -5/ 6, 1/ 6 In two dimensional space the graph of an equation relating the variables x and y is the set of all point (x, y) whose co-ordinates satisfy the equation. Usually, such graphs are curves. In three dimensional space the graph of an equation relating the variables x, y and z is the set of all point (x, y, z) whose co-ordinates satisfy the equation. Usually, such graphs are surfaces. Intersection of two surfaces Intersection of two surfaces is a curve in three dimensional space. It is the reason that a curve in three dimensional space is represented by two equations representing the intersecting surfaces. Intersection of Cone and Sphere Intersection of Two Planes If the two planes are not parallel, then they intersect and their intersection is a straight line. Thus, two non-parallel planes represent a straight line given by two simultaneous linear equations in x, y and z and are known as non-symmetric form of equations of a straight line. 13

14 3-Elements of three dimensional geometry REGION DESCRIPTION EQUATION xy-plane Consists of all points of the form (x, y, ) z = xz-plane Consists of all points of the form (x,, z) y = yz-plane Consists of all points of the form (, y, z) x = x-axis Consists of all points of the form (x,, ) y =, z = y-axis Consists of all points of the form (, y, ) z =, x = z-axis Consists of all points of the form (,, z) x =, y = Planes parallel to Co-ordinate Planes General Equation of Plane Any equation of the form ax + by + cz + d = where a, b, c, d are real numbers,represent a plane. Sphere 14

15 3-Elements of three dimensional geometry Right Circular Cone Horizontal Circular Cylinder 15

16 3-Elements of three dimensional geometry Horizontal Elliptic Cylinder Overview of Lecture # 3 Chapter # 14 Three Diamentional Space Page # 657 Book CALCULUS by HOWARD ANTON 16

17 4-Polar Coordinates Lecture -4 Polar co-ordinates You know that position of any point in the plane can be obtained by the two perpendicular lines known as x and y axis and together we call it as Cartesian coordinates for plane. Beside this coordinate system we have another coordinate system which can also use for obtaining the position of any point in the plane. In that coordinate system we represent position of each particle in the plane by r and θ where r is the distance from a fixed point known as pole and θ is the measure of the angle. P (r, θ) O θ Initial ray O is known as pole. Conversion formula from polar to Cartesian coordinates and vice versa P(x, y) =P(r, θ ) r θ x y From above diagram and remembering the trigonometric ratios we can write x = r cos θ, y = r sin θ. Now squaring these two equations and adding we get, x + y = r Dividing these equations we get y/x = tanθ These two equations gives the relation between the Plane polar and Plane Cartesian coordinates. Rectangular co-ordinates for 3d Since you know that the position of any point in the 3d can be obtained by the three mutually perpendicular lines known as x,y and z axis and also shown in figure below, these coordinate axis are known as Rectangular coordinate system. 17

18 4-Polar Coordinates Cylindrical co-ordinates Beside the Rectangular coordinate system we have another coordinate system which is used for getting the position of the any particle is in space known as the cylindrical coordinate system as shown in the figure below. Spherical co-ordinates Beside the Rectangular and Cylindrical coordinate systems we have another coordinate system which is used for getting the position of the any particle is in space known as the spherical coordinate system as shown in the figure below. 18

19 4-Polar Coordinates Conversion formulas between rectangular and cylindrical co-ordinates Now we will find out the relation between the Rectangular coordinate system and Cylindrical coordinates. For this consider any point in the space and consider the position of this point in both the axis as shown in the figure below. In the figure we have the projection of the point P in the xy-plane and write its position in plane polar coordinates and also represent the angle θ now from that projection we draw perpendicular to both of the axis and using the trigonometric ratios find out the following relations. (r, θ, z) (x, y, z) x = rcosθ, y = rsinθ, z = z r = x +y, tanθ = y x, z = z Conversion formulas between cylindrical and spherical co-ordinates Now we will find out the relation between spherical coordinate system and Cylindrical coordinate system. For this consider any point in the space and consider the position of this point in both the axis as shown in the figure below. First we will find the relation between Planes polar to spherical, from the above figure you can easily see that from the two right angled triangles we have the following relations. (ρ,θ,φ ) (r, θ, z) 19

20 4-Polar Coordinates r = ρ sinφ, θ = θ, z = ρ cos φ Now from these equations we will solve the first and second equation for ρ and φ. Thus we have (r, θ, z) (ρ,θ,φ ) ρ = r +z θ = θ, tan φ = r z Conversion formulas between rectangular and spherical co-ordinates (ρ, θ, Φ) (x, y, z) Since we know that the relation between Cartesian coordinates and Polar coordinates are x = r cos θ, y = r sin θ and z = z.we also know the relation between Spherical and cylindrical coordinates are, r = ρ sinφ, θ = θ, z = ρ cos φ Now putting this value of r and z in the above formulas we get the relation between spherical coordinate system and Cartesian coordinate system. Now we will find ( x, y, z) ( ρ, θ, Φ) x + y +z = (ρsin Φ cos θ) + (ρsin Φ sin θ) + (ρ cos Φ) = ρ {sin Φ(cos θ + sin θ) +cos Φ)} = ρ (sin Φ + cos Φ) = ρ ρ = x + y +z Tanθ = y/x and Cos Φ = z x + y +z Constant surfaces in rectangular co-ordinates The surfaces represented by equations of the form x = x, y = y, z = z where x o, y o, z o are constants, are planes parallel to the xy-plane, xz-plane and xyplane, respectively. Also shown in the figure

21 4-Polar Coordinates Constant surfaces in cylindrical co-ordinates The surface r = ro is a right cylinder of radius ro centered on the z-axis. At each point (r, θ, z) this surface on this cylinder, r has the value r, z is unrestricted and θ <. The surface θ = θ is a half plane attached along the z-axis and making angle θ with the positive x-axis. At each point (r, θ, z) on the surface,θ has the value θ, z is unrestricted and r. The surfaces z = z o is a horizontal plane. At each point (r, θ, z) this surface z has the value z, but r and θ are unrestricted as shown in the figure below. Constant surfaces in spherical co-ordinates The surface ρ = ρ o consists of all points whose distance ρ from origin is ρ o. Assuming that ρ o to be nonnegative, this is a sphere of radius ρ o centered at the origin. The surface θ = θ is a half plane attached along the z-axis and making angle θ with the positive x- axis. The surface Φ = Φ consists of all points from which a line segment to the origin makes an angle of Φ with the positive z-axis. Depending on whether < Φ < / or / < Φ <, this will be a cone opening up or opening down. If Φ = /, then the cone is flat and the surface is the xy-plane. 1

22 4-Polar Coordinates Spherical Co-ordinates in Navigation Spherical co-ordinates are related to longitude and latitude coordinates used in navigation. Let us consider a right handed rectangular coordinate system with origin at earth s center, positive z-axis passing through the north pole,and x-axis passing through the prime meridian. Considering earth to be a perfect sphere of radius ρ = 4 miles, then each point has spherical coordinates of the form (4,θ,Φ) where Φ and θ determine the latitude and longitude of the point. Longitude is specified in degree east or west of the prime meridian and latitudes is specified in degree north or south of the equator. Domain of the Function In the above definitions the set D is the domain of the function. The Set of all values which the function assigns for every element of the domain is called the Range of the function. When the range consist of real numbers the functions are called the real valued function. NATURAL DOMAIN Natural domain consists of all points at which the formula has no divisions by zero and produces only real numbers. Examples Consider the Function ϖ = y x. Then the domain of the function is y x Which can be shown in the plane as

23 4-Polar Coordinates and Range of the function is [, ). Domain of function w = 1/xy is the whole xy- plane Excluding x-axis and y-axis, because at x and y axis all the points has x and y coordinates as and thus the defining formula for the function gives us 1/. So we exclude them. 3

24 5-Limits of multivariable function f (x,y)= sin -1 (x+y) Lecture No-5 Limit of Multivariable Function Domain of f is the region in which -1 x +y 1 x = -1 y-axis x =1-1 x +y 1 y = 1 x-axis y = -1 Domains and Ranges Functions Domain Range ω = x + y + z Entire space [, ) ω = 1 x + y + z (x,y,z) (,, ) (, ) ω = xy lnz Half space z > (, ) Examples of domain of a function f(x, y) = xy y - 1 Domain of f consists of region in xy plane where y 1 f ( x, y ) = x + y - 4 Domain of f consists of region in xy plane where x + y 4 As shown in the figure 4

25 5-Limits of multivariable function f(x, y) = lnxy Domain of f consists of region lying in first and third quadrants in xy plane as shown in above figure right side. f(x, y,z) =e xyz Domain of f consists of region of three dimensional space 4 - x f(x, y) = y + 3 Domain of f consists of region in xy plane x 4,- x x = - y x = f ( x, y, z ) = 5 - x - - y z Domain of f consists of region in three dimensional space occupied by sphere centre at (,, ) and radius 5. f (x, y) = x 3 + x y - xy - y x + y f(,) is not defined but we see that limit exits. 5

26 5-Limits of multivariable function Approaching to (,) through x-axis f (x,y) Approaching to (,) through y-axis f (x,y) (.5,).5 (,.1) -.1 (.5,).65 (,.1) -.1 (.1,).1 (,.1).1 (-.5,).65 (,-.1).1 (-.1,).1 (,-.1).1 Approaching to (,) through y = x f (x,y) (.5,.5) -.5 (.1,.1) -.9 (.1,.1) -.99 (-.5,-.5).75 (-.1,-.1).11 (-.1,-.1).11 6

27 5-Limits of multivariable function Example f (x,y) = xy x +y f(,) is not defined and we see that limit also does not exit. Approaching to (,) through x-axis (y = ) f (x,y) Approaching to (,) through y = x f (x,y) (.5, ) (.5,.5 ).5 (.1, ) (.5,.5 ).5 (.1, ) (.1,.1 ).5 (.1, ) (.5,.5 ).5 (.1, ) (.1,.1 ).5 ( -.5, ) ( -.5,-.5 ).5 ( -.1, ) ( -.5,-.5 ).5 ( -.1, ) ( -.1,-.1 ).5 ( -.1, ) ( -.5,-.5 ).5 ( -.1, ) ( -.1,-.1 ).5 xy lim x + y = (along y = ) (x,y) (,) xy lim x + y =.5 (along y = x) (x,y) (,) xy lim x + y does not exist. (x,y) (,) Example xy lim ( x, y ) (, ) x + y Let ( x, y ) approach (, ) along the line y = x. xy x. x 1 f (x, y ) = x + y = x + x = x = 1 7

28 5-Limits of multivariable function Let (x, y ) approach (, ) along the line y =. f (x, y ) = x. ( ) x + () =, x. Thus f (x, y) assumes two different values as (x,y) approaches (,) along two different paths. lim f (x, y ) does not exist. (x, y) (, ) We can approach a point in space through infinite paths some of them are shown in the y figure below. (x,y ) (xy) x Rule for Non-Existence of a Limit If in Lim ( x, y ) ( a, b ) f ( x, y) We get two or more different values as we approach (a, b) along different paths, then Lim f(x, y) (x, y) (a, b) does not exist. The paths along which (a, b) is approached may be straight lines or plane curves through (a, b). Example Lim (x, y ) (, 1 ) x 3 + x y x y x + y = Lim ( x, y ) (, 1 ) (x 3 + x y xy y ) Lim ( x, y ) (, 1 ) (x + y ) 8

29 5-Limits of multivariable function = Lim ( x, y) (, 1 ) (x 3 + x y x y ) Lim ( x, y) (, 1 ) (x + y ) Example Lim ( x, y ) (,, ) xy x + y We set x = r cos θ, y = r sin θ then x x + y = r cos θ. r sin θ r cos θ + r sin θ = r cos θ sin θ, for r > Since r = x + y, r as ( x, y ) (, ), Lim ( x, y ) (, ) RULES FOR LIMIT x x + y = Lim r r cos θ sin θ =, since cos θ sin θ 1 for all value of θ. If (a) (b) lim f ( x, y) = L and lim g( x, y) = L Then 1 (, ) (, ) (, ) (, ) xy x y xy x y lim cf ( x, y) = cl (if c is constant) ( xy, ) ( x, y) lim { f ( xy, ) + gxy (, )} = L+ L ( xy, ) ( x, y) 1 1 (d) (e) lim { f ( xy, ) gxy (, )} = L L ( xy, ) ( x, y) lim { f ( xygxy, ) (, )} = LL ( xy, ) ( x, y) lim ( xy, ) ( x, y) f( x, y) gxy (, ) L L 1 1 = (if L = ) lim ( xy, ) ( x, y) 1 c= c (c a constant), lim ( xy, ) ( x, y) x = x, lim ( xy, ) ( x, y) y = y Similarly for the function of three variables. 9

30 5-Limits of multivariable function Overview of lecture# 5 In this lecture we recall you all the limit concept which are prerequisite for this course and you can find all these concepts in the chapter # 16 (topic # 16.)of your Calculus By Howard Anton. 3

31 6-Geometry of continuous functions Lecture No -6 Geometry of continuous functions Geometry of continuous functions in one variable or Informal definition of continuity of function of one variable. A function is continuous if we draw its graph by a pen then the pen is not raised so that there is no gap in the graph of the function Geometry of continuous functions in two variables or Informal definition of continuity of function of two variables. The graph of a continuous function of two variables to be constructed from a thin sheet of clay that has been hollowed and pinched into peaks and valleys without creating tears or pinholes. Continuity of functions of two variables A function f of two variables is called continuous at the point (x,y ) if 1. f (x,y ) if defined.. 3. lim f ( xy, ) exists. ( xy, ) ( x, y) lim f ( xy, ) ( xy, ) ( x, y) = f (x,y ). The requirement that f (x,y ) must be defined at the point (x,y ) eliminates the possibility of a hole in the surface z = f (x,y ) above the point (x,y ). Justification of three points involving in the definition of continuity. (1) Consider the function of two variables x + y ln( x + y ) now as we know that the Log function is not defined at, it means that when x = and y = our function x + y ln( x + y ) is not defined. Consequently the surface z = x + y ln( x + y ) will have a hole just above the point (,)as shown in the graph of x + y ln( x + y ) () The requirement that lim f ( xy, ) exists ensures us that the surface z = f(x,y) of ( xy, ) ( x, y) the function f(x,y) doesn t become infinite at (x,y ) or doesn t oscillate widely. 31

32 6-Geometry of continuous functions 1 Consider the function of two variables now as we know that the Natural domain x + y of the function is whole the plane except origin. Because at origin we have x = and y = in the defining formula of the function we will have at that point 1/ which is infinity. 1 Thus the limit of the function does not exists at origin. Consequently the x + y 1 surface z = will approaches towards infinity when we approaches towards x + y origin as shown in the figure above. (3) The requirement that lim f ( xy, ) ( xy, ) ( x, y) = f (x,y ) ensures us that the surface z = f(x,y) of the function f(x,y) doesn t have a vertical jump or step above the point (x,y ). Consider the function of two variables if x and y f( x, y) = 1 otherwise now as we know that the Natural domain of the function is whole the plane. But you should note that the function has one value for all the points in the plane for which both x and y have nonnegative values. And value 1 for all other points in the plane. Consequently the surface if x and y z = f( x, y) = has a jump as shown in the figure 1 otherwise 3

33 6-Geometry of continuous functions Example Check whether the limit exists or not for the function x lim f( x, y) = x + y ( xy, ) (,) Solution: First we will calculate the Limit of the function along x-axis and we get lim (, ) x f x y = = 1 (Along x-axis) ( xy, ) (,) x + Now we will find out the limit of the function along y-axis and we note that the limit y is lim f( x, y) = = 1(Along y-axis). Now we will find out the limit of the ( xy, ) (,) y + x 1 function along the line y = x and we note that lim f( x, y) = = (Along y = x) ( xy, ) (,) x + x It means that limit of the function at (,) doesn t exists because it has different values along different paths. Thus the function cannot be continuous at (,). And also note that the function id not defined at (,) and hence it doesn t satisfy two conditions of the continuity. Example Check the continuity of the function at (,) f x, y) = sin( x x + y 1 + y ) ( if ( x, y) (,) if ( x, y) = (,) Solution: First we will note that the function is defined on the point where we have to check the Continuity that is the function has value at (,). Next we will find out the sin x Limit of the function at (,) and in evaluating this limit we use the result lim = 1 x x and note that 33

34 6-Geometry of continuous functions Sin(x + y ) lim f(x,y) = lim x + y (x,y) (,) (x,y) (,) =1 = f(, ) This shows that f is continuous at (,) CONTINUITY OF FUNCTION OF THREE VARIABLES A function f of three variables is called continuous at a point (x,y,z ) if 1. f (x,y,z ) if defined.. lim f ( xyz,, ) exists; 3 ( xyz,, ) ( x, y, z) lim f ( xyz,, ) = f(x,y,z ). ( xyz,, ) ( x, y, z) EXAMPLE Check the continuity of the function y + 1 f( x, y, z) = x + y 1 Solution: First of all note that the given function is not defined on the cylinder x + y 1=.Thus the function is not continuous on the cylinder x + y 1= And continuous at all other points of its domain. RULES FOR CONTINOUS FUNCTIONS (a) If g and h are continuous functions of one variable, then f(x, y) = g(x)h(y) is a continuous function of x and y (b) If g is a continuous function of one variable and h is a continuous function of two variables, then their composition f(x, y) = g(h(x,y)) is a continuous function of x and y. A composition of continuous functions is continuous. A sum, difference, or product of continuous functions is continuous. A quotient of continuous function is continuous, expect where the denominator is zero. EXAMPLE OF PRODUCT OF FUNCTIONS TO BE CONTINUED In general, any function of the form f(x, y) = Ax m y n (m and n non negative integers) is continuous because it is the product of the continuous functions Ax m and y n. The function f(x, y) = 3x y 5 is continuous because it is the product of the continuous functions g(x) = 3x and h(y) = y 5. CONTINUOUS EVERYWHERE A function f that is continuous at each point of a region R in -dimensional space or 3-dimensional space is said to be continuous on R. A function that is continuous at every point in -dimensional space or 3-dimensional space is called continuous everywhere or simply continuous. 34

35 6-Geometry of continuous functions Example: (1)f (x,y) = ln(x y +1) The function f is continuous in the whole region where x > y-1, y < x+1.and its region is shown in figure below. y < x+1 1 f ( xy, ) = e xy () The function f is continuous in the whole region of xy-plane. 1 (3) f ( xy, ) = tan ( y x) The function f is continuous in the whole region of xy- plane. (4) f ( xy, ) = y x The function is continuous where x y y x y Partial derivative Let f be a function of x and y. If we hold y constant, say y = y and view x as a variable, then f(x, y ) is a function of x-alone. If this function is differentiable at x = x o, then the value of this derivative is denoted by f x (x o, y ) and is called the Partial derivative of f with respect of x at the point (x o, y ). Similarly, if we hold x constant, say x = x and view y as a variable, then f (x, y ) is a function of y alone. If this function is differentiable at y = y, then the value of this derivative is denoted by f y (x, y ) and is called the Partial derivative of f with respect of y at the point (x, y ) 35

36 6-Geometry of continuous functions Example 3 f ( xy, ) = xy + y+ 4x Treating y as a constant and differentiating with respect to x, we obtain f x (x, y) = 6x y + 4 Treating x as a constant and differentiating with respect to y, we obtain f y (x, y) = 4x 3 y + Substituting x = 1 and y = in these partial-derivative formulas yields. f x (1, ) = 6(1) () + 4 = 8 f y (1, ) = 4(1) 3 () + = 1 Example Example Z = 4x - y + 7x 4 y 5 z 3 5 = 8x + 8x y x z 4 4 = + 35x y y z = f( x, y) = x sin y Then to find the derivative of f with respect to x we treat y as a constant therefore z = f x = xsin y x Then to find the derivative of f with respect to y we treat x as a constant therefore z = f y = x sin ycos y y = x sin y Example z ln x + y = x + y By using the properties of the ln we can write it as z = ln(x + y ) ln (x + y) z x = 1 1 x + y. x x + y = x + xy x y (x + y )(x + y) = x + xy y (x + y )(x + y) Similarly, (or by symmetry) z y = y + xy x (x + y )(x + y) 36

37 6-Geometry of continuous functions Example 4 3 z = x sin( xy ) z = 4 3 x sin( xy ) x x = x sin( xy ) + sin( xy ) ( x ) x x = x cos( xy ) y + sin( xy )4x z = x y cos( xy ) + sin( xy ) x z = 4 3 x sin( xy ) y y = x sin( xy ) + sin( xy ) ( x ) y y = x cos( xy ).3xy + sin( xy ). = cos( ) xy xy Example z = cos(x 5 y 4 ) z = sin( x y ) ( x y ) x x = 5x y sin( x y ) z = sin( x y ) ( x y ) y y = 4x y sin( x y ) Example w = x yz x w = 6y - xz dy dw = 8z - xy dz w = x +3y +4z -xyz 37

38 7-Geometric meaning of partial derivative Lecture No -7 Geometric meaning of partial derivative Geometric meaning of partial derivative z = f(x, y) Partial derivative of f with respect of x is denoted by z x or f x or f x Partial derivative of f with respect of y is denoted by z f o r fy or y y, Partial Derivatives Let z = f(x, y) be a function of two variable defined on a certain domain D. For a given change x in x, keeping y as it is, the change z in z, is given by If the ratio z = f (x + x, y) f (x, y) Δ z Δ x = f(x + Δ x, y) f(x, y) Δ x approaches to a finite limit as x, then this limit is called Partial derivative of f with respect of x. Similarly for a given change y in y, keeping x as it is, the change z in z, is given by z = f (x, y + y) f (x, y) If the ratio Δz Δy = f(x, y+ Δy) f(x, y) Δy approaches to a finite limit as y, then this limit is called Partial derivative of f with respect of y. Geometric Meaning of Partial Derivatives Suppose z = f ( x, y ) is a function of two variables.the graph of f is a surface. Let P be a point on the graph with coordinates ( xo, y o, f ( x o, yo )). If a point starting from P, changes its position on the surface such that y remains constant, then the locus of this point is the curve of intersection of z= f (x, y ) and y = constant.on this curve, z x is derivative of z = f ( x, y ) with respect tox with y constant. Thus z = slope of the tangent to this curve at P x 38

39 7-Geometric meaning of partial derivative z Similarly, is the gradient of the y tangent at P to the curve of intersection of z = f (x, y ) and x = constant. As shown in the figure below (left) Also together these tangent lines are shown in figure below (right). Partial Derivatives of Higher Orders The partial derivatives f x and f y of a function f of two variables x and y, being functions of x and y, may possess derivati ves. In such cases, the second order partial derivatives are defined as below. Example f = y f = y f y = y (f y ) = (f y ) y = f yy = f y Thus, there are four second order partial derivatives for a function z = f( x, y ). The partial derivatives f xy and f yx are called mixed second partials and are not equal in general. Partial derivatives of order more than two can be defined in a similar manner. z = arc sin z x = 1 z y = 1 f x x f = y x x y 1 x y 1 x y z y x = y x x y.. 1 y = 1 y x. x y = z = f x = x (f x ) = (f x ) x = f xx = f x f y x = y (f x ) = (f x ) y = f xy f x y = x (f y ) = (f y ) x = f yx x y y x = 1 (y x ) 3/. y = y (y x ) 3/ 39

40 7-Geometric meaning of partial derivative z x y = z x y 1 = y y x x y x (y x ) 3/ Hence = y + x x y(y x ) 3/ = z x y = z y x y (y x ) 3/ Example f ( x, y ) = x cos y + Laplace s Equation f = cos y + ye x x f f = = sin y x y x f f x = = ye x x x For a function w = f(x,y,z) y + f(x, y) = x cosy + y e x f = x siny + e x y f = siny + e x x y f y = f = -x cos y y y The equation f x + f f y z = is called Laplace s equation. x ye e x 4

41 7-Geometric meaning of partial derivative Example f ( x, y ) = e x sin y + e y cos x, f = e x sin y e y sin x x f x = e x sin y e y cos x f y = ex cosy + e y cosx f y = -e x siny +e y cosx Adding both partial second order derivative, we have f x + f y = e x sin y e y cos x e x sin y + e y cos x = Euler s theorem The mixed derivative theorem If f(x,y) and its partial derivatives f x, f y, f xy and f yx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then Advantage of Euler s theorem f xy (a, b) = f yx (a, b) y e w = xy + y +1 The symbol w/ x y tell us to differentiate first with respect to y and then with respect to x. However, if we postpone the differentiation with respect to y and differentiate first with respect to x, we get the answer more quickly. w = x y w y x = 1 Overview of lecture# 7 Chapter # 16 Partial derivatives Page # 79 Article #

42 8-Euler theorem chain rule 4 Lecture No- 8 More About Euler Theorem Chain Rule Order of differentiation For a function If we are interested to find, that is, differentiating in the order firstly w.r.t. x and then w.r.t. y, calculation will involve many steps making the job difficult. But if we differentiate this function with respect to y, firstly and then with respect to x secondly then the value of this fifth order derivative can be calculated in a few steps. EXAMPLE In general, the order of differentiation in an nth order partial derivatice can be change without affecting the final result whenever the function and all its partial derivatives of order n are continuous. For example, if f and its partial order derivatives of the first, second, and third orders are continuous on an open set, then at each point of the set, yyx yxy xyy f f f = = or in another notation y x f y x y f x y f = = y x y x y x f + = ), ( ) ( ) ( ) ( ) ( ) ( ), ( y x y x x y x y x x y x y x f x + + = ) ( )(1) ( )(1) ( y x y x y x + = ) ( y x y = f (x,y) = y x 4 e x + 5 f y 3 x 3 5 = y x f ) ( ) ( ) ( ) ( ) ( ), ( y x y x y y x y x y y x y x f y + + = ) ( ) 1 )( ( ) 1 )( ( y x y x y x + = ) ( y x x =

43 8-Euler theorem chain rule EXAMPLE f f x y f ( x, EXAMPLE y ) = ( x, y) = 3 x ( x, y) = x x y e 3 e y e 3 y + + y y sec y 3 sec x sec x tan x 1 x x f f ( x, y) = xy x( x, y) = xye + f x x xy ye xy x y e = xye xy ( + xy) (1)(1) (1,1) = (1)(1) e [ + (1)(1)] = 3e f ( x, y) = x xy ye f xy y( x, y) = x e + xy f y x 3 ye = x e (1 + xy) (1)(1) (1,1) = (1)(1) e [1 + (1)(1)] = e xy Example f x f ( x, y) = x cos( xy) f x ( x, y) = xcos( xy) x ysin( xy) (1, ) = (1 )cos( ) (1 ) ( )sin( = 4 3 f y ( x, y) = x sin( xy) 3 f ( 1, ) = ( 1 ) sin( y = 1 8 ) ) 43

44 8-Euler theorem chain rule EXAMPLE w = ( 4 x 3y + z) w 4 = ( 4 x 3 y + z ) x w 3 = 4 ( 4 x 3 y + z ) y x 3 w = 144(4x 3y+z) z y x 4 w = 576 ( 4 x 3 y + z ) z y x 5 Chain Rule in function of One variable Given that w= f(x) and x = g(t), we find From w = f(x), we get From x = g(t), we get Then Example dw dt = dw dx dw dx dx dt d x dt dw dt as follows: w = x + 4, x = Sint By Substitution w = Sint + 4 dw dt = Cost w = x + 4 dw dx = 1 x = Sint dx dt = Cost By Chain Rule dw dt = dw dx = 1. Cost = Cost dt dt 44

45 8-Euler theorem chain rule Chain rule in function of one variable y is a function of u, u is a function of v v is a function of w, w is a function of z z is a function of x. Ultimately y is function of x so we can talk about dy dx and by chain rule it is given by dy dx = dy du du dv dv dw dw dz dz dx w = f(x,y), x = g(t), y = f(t) Dependent variable w x w = f(x,y) w y Intermediate variables dx dt x dy dt y dw w dx w dy = + dt x dt y dt t Independent variables EXAMPLE BY SUBSTITUTION w = xy x = cost, y = sint w = cost sint = 1 sint cot = 1 sint dw dt = 1 cost. = cos t 45

46 8-Euler theorem chain rule EXAMPLE w y dx dt = sint, w= xy, x= cos t, and y = sin t w = x = y x dy dt = cost, dw dt = w x dx dt + w y dy dt = (sin t)( sin t) + (cos t)(cos t) = - sin t + cos t = cos t EXAMPLE z = 3x y 3 x = t 4, y = t z x = z 6xy3, y = 9 x y dx = 4t 3 dy, = t dt dt dz dt = z dx + z dy x dt y dt = (6xy 3 ) (4t 3 ) + 9x y (t) = 6 (t 4 ) (t 6 ) (4t 3 ) + 9 (t 8 ) (t 4 ) (t) = 4 t t 13 = 4t 13 EXAMPLE 4 z = 1 + x - xy x = ln y = t z x = x - xy z y = x - xy dx dt = 1 t, d dt =.(- 8xy 3 ) = 3-4xy x - xy 46

47 8-Euler theorem chain rule dz dt = z x EXAMPLE dx dt + z y. dy dt 1 y = 1 + x - xy 4 t t - 4xy 1 + x - xy = 1 + x - xy 4 4-4xy 3 t 1 1-t 4 = - 4 (lnt) t lint - ( lnt ) t 4 t 1 1 = 1 + lnt - t 4 lnt t - t3-4t 3 lnt z = ln (x + y) x = t, y = t /3 z x = 1 x + y. 4x = 4 x + y z y = 1 x + y, dx = 1 dt 1t, dy dt w = f(x,y,z), x = g(t),y = f(t), z = h(t) = 3 t -1/3 w x w = f(x,y,z) w y Dependent variable w z y z dx dt dy dt dz dt Independent variables t Overview of Lecture#8 Chapter # 16 Topic # 16.4 Page # 799 Book Calculus By Haward Anton 47

48 8-Euler theorem chain rule Lecture No - 9 Examples First of all we revise the example which we did in our 8 th lecture. Consider w = f(x,y,z) Where x = g(t), y = f(t), z = h(t) Then dw w dx w dy w dz = + + dt x dt y dt z dt Example: w = x + y + z + 4 x = e t, y = cost, z = t + 4 w x = x, w y = 1, w z = 1 dx = e t dy, dt dt = Sint, dz = 1 dt dw = w dx + w dy + w dt x dt y dt z. dz dt = (x) (e t ) + (1). ( Sint) + (1) (1) = (e t ) (e t ) Sint + 1 = e t Sint + 1 Consider w = f(x), where x = g(r, s). Now it is clear from the figure that x is intermediate variable and we can write. w dw x = r dx r Example: and w dw x = s dx s w = Sin x + x, x = 3r + 4s dw = Cosx + x dx x r = 3 x s = 4 w r = dw dx. x r = (Cosx + x). 3 = 3 Cos (3r+4s) + 6 (3r + 4s) = 3 Cos (3r + 4s) + 18r + 4s w s = dw dx. x s = (Cosx + x). 4 = 4 Cosx + 8x = 4 Cos (3r + 4s) + 8 (3r + 4s) = 4 Cos (3r + 4s) + 4r + 3s Dependent variable x r w = f(x) dw dx x Intermediate variables x s s 48

49 9-Examples Consider the function w = f(x,y), Where x = g(r, s), y = h(r, s) w = f(x,y) Dependent variable w x w y x r x x s y r y y s Intermediate variables r s r r w r = w x x r + w y y r Similarly if you differentiate the function w with respect to s we will get And we have w s = w x x s + w y y s 49

50 9-Examples Consider the function w = f(x,y,z), Where x = g(r, s), y = h(r,s), z = k(r, s) w = f(x,y,z) Dependent variable r x r x z intermediate variables y x s s y r r y s p r z r z s s Independent variables Thus we have w r = w x Similarly if we differentiate with respect to s then we have, w s = w x x r + w y y r + w z z r x s + w y y s + w z z s Example: Consider the function w First we will calculate x y z = + +, x =, r 1 = (1) () ( z)() + + s s r = s s r s = + ln, z = r y r s w w = 1 = w x 1 y z = z = = r = x y z r s r r w w x w y w z Now as we know that = + + By putting the values from above we get r x r y r z r δ w 1 = (1) + ()( r) + ( z)() δ r s 1 1 = + 4 r + (4 r )() = + 1r Now s s x r y 1 z = = = s s s s s So we can calculate w w x w y w z = + + s x r y r z r 5

51 9-Examples Remembering the different Forms of the chain rule: The best thing to do is to draw appropriate tree diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To find the derivative of dependent variable with respect to the selected independent variable, start at the dependent variable and read down each branch of the tree to the independent variable, calculating and multiplying the derivatives along the branch. Then add the products you found for the different branches. ω p Example: The Chain Rule for Functions of Many Variables Suppose ω = f (x, y,., υ) is a differentiable function of the variab les x, y,.., υ (a fin ite set) a nd the x, y,, υ are differentiable functions of p, q,, t (ano ther finite set). Then ω is a differentiable function of the variab les p thro ugh t a nd the partial derivatives of ω with respect to these variables are given by equations of the form = ω x x p + ω y y p The other equations are obtained by replacing p by q,, t, one at a time. One way to remember last equation is to think of the right-hand side as the dot product of two vectors with components. ω x, ω ω and y υ Derivatives of ω with respect to the intermedaite variables + + ω υ υ p. x p, y υ p p Derivatives of the intermedaite variables with respect to the selected independent variable r s t u w= ln( e + e + e + e ) Taking ln of both sides of the given equation we get w r s t u e = e + e + e + e Now Taking partial derivative with respect to r, s, u, and t we get ew= e w= e w r r w r r w s s w, ew e w e w u u w = =, ewu = e wu = e and s s ew= e w= e w t t w t t 51

52 9-Examples Now since we have wr = r w e Now Differentiate it partially w.r.t. s r w w = e ( w ) rs wrs = e = r w e r s w e + s s w (Here we use the value of w s ) Now differentiate it partially w.r.t. t and using the value of r+ s w wrst = e ( wt ) = e r+ s w e t w 3 w = e r+ s+ t w rst Now differentiate it partially w.r.t. u we get, r+ s 3w wrstu e ( 3 wu ) get, r+ s+ t 3w u w wrstu = e e r+ s+ t+ u 4w wrstu = 6e w t we get, = and by putting the value of u 6 ( ) w, we 5

53 11-The triple scalar or Box product Lecture No -1 Introduction to vectors Some of things we measure are determined by their magnitude. But some times we need magnitude as well as direction to describe the quantities. For Example, To describe a force, We need direction in which that force is acting (Direction) as well as how large it is (Magnitude). Other Example is the body s Velocity; we have to know where the body is headed as well as how fast it is. Quantities that have direction as well as magnitude are usually represented by arrows that point the direction of the action and whose lengths give magnitude of the action in term of a suitably chosen unit. A vector in the plane is a directed line segment. B v r A r uuur v= AB Vectors are usually described by the single bold face roman letters or letter with an arrow. The vector defined by the directed line segment from point A to point B is written as AB. Magnitude or Length Of a Vector : Magnitude of the vector v r is denoted by v r = AB is the length of the line segment AB Unit vector Any Vector whose Magnitude or length is 1 is a unit vector. Unit vector in the direction of vector v r is denoted by v $. and is given by r v v$ = v Addition Of Vectors B r O a r uuur This diagram shows three vectors, in two vectors one vector OA is connected with tail of uuur uuur vector AB. The tail of third vector OB is connected with the tail of OA and head is uuur connected with the head of vector AB.This third vector is called Resultant vector. b r A 53

54 11-The triple scalar or Box product The resultant vector can be written as r = a + b Similarly r = a + b + c + d + e + f f E e D d C F c r O a A b B Equal vectors Two vectors are equal or same vectors if they have same magnitude and direction. a Opposite vectors opposite direction. a Two vectors are opposite vectors if they have same magnitude and a -a Parallel vectors Two vector are parallel if one vector is scalar multiple of the other. b = λa where λ is a non zero scalar. 54

55 11-The triple scalar or Box product r = x i+ y j + z k Addition and subtraction of two vectors in rectangular component: Let a = a 1 i + a j + a 3 k and b = b 1 i + b j + b 3 k a + b = (a 1 i + a j + a 3 k) + (b 1 i + b j + b 3 k) = (a 1 + b 1 )i + (a + b )j + (a 3 + b 3 )k a - b = (a 1 i + a j + a 3 k) - (b 1 i + b j + b 3 k) = (a 1 - b 1 )i + (a - b )j + (a 3 - b 3 )k I th component of first vector is added (subtracted) to the ith component of second vector, jth component of first vector is added (subtracted) to the jth component of second vector, similarly kth component of first vector is added (subtracted) to the kth component of second vector, Multiplication of a vector by a scalar a a 3a -a Any vector a can be written as a = a aˆ Scalar product Scalar product (dot product) ( a dot b ) of vector a and b is the number a.b = a b cos θ. where θ is the angle between a and b. In word, a.b is the length of a times the length of b times the cosine of the angle between a and b. Remark:- a.b = b.a 55

56 11-The triple scalar or Box product This is known as commutative law. Some Results of Scalar Product a.b = a b cos θ. 1. a b If This means that if a is perpendicular to b. Then a.b= Also i.j= = j.i j.k= =k.j k.i= =i.k. If a b That means a is parallel to b. Then a.b = a b if we replace b by a then a.a = a a a.a = a a = a.a so i.i=j.j=k.k=1 Example If a = 3 k and b = i + k, then a. = a b cos θ = (3) () cos 4 = 6. = 3. 56

57 11-The triple scalar or Box product EXPRESSION FOR a.b IN COMPONENT FORM a = a 1 i + a j + a 3 k and b = b 1 i + b j + b 3 k a.b = (a 1 i + a j + a 3 k ). (b 1 i + b j + b 3 k ) = a 1 i. (b 1 i + b j + b 3 k ) + a j. (b 1 i + b j + b 3 k ) + a 3 k. (b 1 i + b j + b 3 k ) = a 1 b 1 i. i + a 1 b i.j + a 1 b 3 i. k + a b j. i + a b j. j + a b 3 j. k + a 3 b 1 k. i + + a 3 b k. j + a 3 b 3 k. k = a 1 b 1 (1) + a 1 b () + a 1 b 3 () + a b 1 () + a b (1) + a b 3 () + a 3 b 1 () + + a 3 b () + a 3 b 3 (1) = a 1 b 1 + a b + a 3 b 3 In dot product ith component of vector a will multiply with ith component of vector b, jth component of vector a will multiply with jth component of vector b and kth component of vector a will multiply with kth component of vector b Angle Between two vectors The angle between two nonzero vectors a and b is θ = cos - 1 a. b a b Since the values of the arc lie in [, i ], above equation automatically gives the angle made by a and b. Example a = i j k and b = 6i + 3j + k. a.b = (1)(6) + ( )(3) + ( ) () = = 4 a = (1) + ( ) + ( ) = 9 = 3 b = ( 6 ) + ( 3 ) + ( ) = 49 = 7 θ = cos - 1 a.b a b = cos = cos -1 (3) (7) rad Perpendicular (Orthogonal) Vectors a and b are perpendicular if and only if a.b =. This has two parts If a and b are per perpendicular then a.b=. And if a.b= a and b will be perpendicular. 57

58 11-The triple scalar or Box product Vector Projectio Consider the Projection of a vector b on a vector a making an angle θ with each other B b O θ a C A From right angle triangle OCB COS θ =Base / hypotenuse COS θ = OC b OC = b Cosθ b a Cos θ = a = b.a a = b. a a = b. a a The number b cos θ is called the scalar component of B in the direction of a. s ince b cos θ = B. a, a we can find the scalar component by dotting b with the direction of a Example Vector Projection of b = 6 i + 3 j + k onto a = i - j k is proj a b = b. a a. a a = ( i j k ) = 4 9 ( i j k ) = 4 9 i j k. 58

59 11-The triple scalar or Box product he scalar of component b in T the direction of a b cos θ = b. a a = (6 i + 3 j + k ). 1 3 i 3 j 3 k = 4 3 = 4 3. The C ross Product of Two Vectors in Space Consider t wo nonzero vectors a and b in space. T he vec t or product a b ( a cross b ) to be the vector a b = ( a b sin θ ) n where n is a vector determined by right hand rule. Right-hand rule We start with two nonzero nonparallel vectors A and B.We select a unit vector n Perpendicular to the plane by the right handed rule. This means we choose n to be the unit vector that points the way your right thumb points when your fingers curl through the angle from A to B. The vector A*B is orthogonal to both A and B. Some Results of Cross Product 59

60 11-The triple scalar or Box product a b = a b Sin θ ^n If a b then a b = So a a = i i = j j = k k = If a b then a b = a b ^n i j = k, j i = k j k = i, k j = i k i = j i k = j N ote that this product is not commutative. j i k The Area of a Parallelogram Because n is a unit the ma gnitude of a b is a b = a b sin θ n = A b sin θ This is the area of the determined by a and b a being the base of the and b sin θ the 6

61 11-The triple scalar or Box product a b from the components of a and b Suppose a = a 1 i + a j + a 3k, b = b 1 i + b j + b 3k. a b = (a 1 i + a j + a 3 k ) (b 1 i + b j + b 3k ) = a 1 b 1 i i + a 1 b i j + a 1b 3i k + a b 1j i + a b j j + a b 3j k + a 3 b 1k i + a 3 b k j + a 3b 3k k a = = (a a b 13 i + a 3 b j ) + i a 3 (a k, 1 b 3 a 3b 1) j b = + b (a 11 i b + b a j b + 1 ) b k 3. k. i j k a b = a 1 a a 3 b 1 b b 3 Example: Let a = i + j + k and b = 4i + 3j + k. Then i j k a b= a b= i(1 3) j( + 4) + k(6 + 4) a b= i 6j+ 1k is the required cross product of a and b. Over view of Lecture # 1 Chapter# 14 Article # 14.3, 14.4 Page #

62 11-The triple scalar or Box product Lecture No -11 The Triple Scalar or Box Product The product (a b). c is called the triple scalar product of a, b, and c (in that order). As (a b).c = a b c cos θ the absolute value of the product is the volume of the parallelepiped (parallelogram-sided box) determin-ed by a,b and c By treating the planes of b and c and of c and a as he base planes of the parallelepiped determined by a, b and c we see that (a * b).c = (b c).a=(c a).b Since the dot product is commutative, ( a b).c =a.(b c) a = a 1 i + a j + a 3 k, b = b 1 i + b j + b 3 k. c = c 1 i + c j + c 3 k. i j k a 1 a a 3 a. (b c) = a. b 1 b b 3 b b = a. 3 b 1 b 3 b 1 b i - j + c c 3 c 1 c 3 c 1 c k b b = a 3 b 1 b 1 a 3 b 1 b c c 3 + a c 1 c 3 3 c 1 c = a 1 a a 3 b 1 b b 3 c 1 c c 3 Example a = i + j k, b = i + 3 k, c = 7 j 4 k. 1 1 a.( b c ) = = = = 3 Th e volume is When we solve a.(b c) a then.( b answer c ) = is if we get negative value then Absolute value make it positive and also volume is always positive. 6

63 11-The triple scalar or Box product Gradient of a Scalar Function is called del operator del operator is a vector quantity. Grad means gradient. Gradient is also vector quantity. φ is vector and φ is scalar quantity, Every component of φ will operate with the. Directional Derivative i x + j y + k z, is called del operator Gradient φ is a vector operator defined gradiant as φ is a vector operator defined as grand φ = i x + j grad φ = i y + k x + j y + k z φ = z φ φ, = φ, i x + j y + k z, If f(x,y) is differentiable at (x,y), and if u = (u 1, u ) is a unit vector, then the directional derivative of f at (x, y ) in the direction of u is defined by D uf(x,y ) = f x(x,y )u1 + f y (x,y)u It should be kept in mind that there are infinitely many directional derivatives of z = f(x,y) at a point (x,y ), one for each possible choice of the direction vector u Remarks (Geometrical interpretation) The directional derivative D u f(x,y ) can be interpreted algebraically as the instantaneous rate of change in the direction of u at (x,y ) of z=f(x,y) with respect to the distance parameter s described above, or geometrically as the rise over the run of the tangent line to the curve C at the point Q Example 63

64 11-The triple scalar or Box product T he directional derivative of f(x,y) = 3x y at the point (1, ) in the direction of the vector a = 3i + 4j. f(x,y ) = 3x y f x (x,y) = 6xy, f y (x,y) = 3x so that f x (1, ) = 1, f y (1,) = 3 a = 3 i +4j ^ a = a 1 = (3i + 4j ) Note: a 5 Formula for the directional = 3 5 i j derivative can be written in the following compact form using the D u f ( 1,) = 1 gradient notation D u f(x, y) = f(x, y). ^u = 48 The dot product of the gradient of f 5 with a unit vector ^u produces the f x means that function f(x,y) is differentiating partially with respect to x and f y means that function f(x,y) is differentiating partially with respect to y. Example f(x, y) = x Remarks u = 3 i 4 j + y, P ( 1, 1) u = 3 + ( 4 ) = 5 ^ u = 3 5 i 4 5 j f x = 4x f x ( 1, 1) = 4 f y = y f y ( 1, 1) = D u f (- 1, 1 ) = f x (- 1, 1 )u 1 + f y (- 1, 1 )u = = 4 Another example, In this example we have to find directional derivative of the function f ( xy, ) = x + y at the point P (-1,1) in the direction of u = 3i 4j. To find the directional derivative we again use the above formula If u = u 1 i + u jis a unit vector making an angle θ with the positive x - axis, then u 1 = cos θ and u = sin θ D u f(x,y )=f x (x,y )u 1 + f y (x,y )u can be written in the form D u f(x,y ) = f x (x,y ) cos θ + f y (x,y ) sin θ Example 64

65 11-The triple scalar or Box product T he d irectio na l de riva tive o f e xy at (,) in the direction of the unit vec tor u that makes an angle of /3 w ith the positive x-axis. f(x,y) = e xy f x (x.y) = ye xy, f y (x, y) = x e xy fx (,) =, fy (,) = D u f(, ) = f x (, ) cos 3 + f y(, ) sin 3 = 1 + ( ) 3 = 3 Gradient of function Directional Derivative If f is a function of x and then the gradient of is defined f (x, y) = f x (x, y) i + f y (x, y) j Formula for the directional derivative can be written in the using the gradient D u f(x, y) = f(x, y). ^u following compact form The dot product of the gradient with a unit ^u produces directional derivative of f in direction u. f EXAMPLE f(x,y) = xy 3y, P (5,5) u = 4i + 3j u = = 5 ^u = u = 4 u 5 i j f x = y, fy = x by f x (5, 5 ) = 1, fy (5,5) = f = 1i j D uf(5,5) = f. ^u 4 = = 4,In this example we have to find directional derivative of the function f ( xy, ) = xy 3y at the point P(5,5) in the direction of u = 4i + 3j. To find the directional derivative we again use the above formula 65

66 11-The triple scalar or Box product EXAMPLE Directional derivative of the function f(x, y) = xe y +cos (xy) at the point (,) in the direction ofa = 3 i 4j. ^ u = a a = 3 5 i 4 5 j. f x ( x,y ) = e y y sin (xy) f y (x,y) = xe y xsin(xy) The partial derivatives of f at (, ) are f x (,) = e = 1 f y (,) = e. = The gradient of f at (, ) f (, = x (, ) i + y (, ) j = i + j The derivative of f at ) in the direction of a is ( u f) (, = f (,. = j) 3 5 i 4 5 j = Properties of Directional Derivatives D uf = f. u ^ = f cos θ 1. The function f increases most rapidly when cos θ = 1, or when u is the direction of f. That is, at each point P in its domain, f increases most rapidly ^ in the direction of the gradient vector f at P. The derivative in this direction is D u f = f cos() = f.. Similarly, f decreases most rapidly in the direction of f. The derivative in this direction is D u f= f cos ( ) = f. 3. Any direction u ^ orthogonal of the gradient is a direction of zero change in f because se θ then equals / and D u f= f cos ( /)= f.= 66

67 11-The triple scalar or Box product f (x, y) = x + y a) The direction of rapid change. The function increases most rapidly in the direction of f at (1,1). The gradient is ( f) (1,1) = (x i + yj ) (1,1) = i + j. its direction is ^ u = i + j i + j i + j = ( 1) + ( 1) = 1 i + 1 j. (1,1) are the directions orthogonal to f. b) The directions of zero change The directions of zero change at ^ n = i + j 1 1 and ^ n = 1 i 1 j. 67

68 11-The triple scalar or Box product 68

69 1-Tangent planes to the surfaces Lecture No -1 Tangent planes to the surfaces Normal line to the surfaces If C is a smooth parametric curve on three dimension, then tangent line to C at the point P is the line through P along the unit tangent vector to the C at the P.The concept of a tangent plane builds on this definition. If P(x,y,z) is a point on the Surface S, and if the tangent lines at P to all the smooth curves that pass through P and lies on the surface S all lie in a common plane, then we shall regards that plane to be the tangent plane to the surface S at P. Its normal (the straight line through P and perpendicular to the tangent) is called the surface normal of S at P. Different forms of equation of straight line in two dimensional space 1. Slope intercept form of the Equation of a line. y = mx + c Where m is the slope and c is y intercept..point_slope Form Let m be the slope and P( x, y) be the point of required line, then y y = m (x x ) 3. General Equation of straight line Ax + By + C = m = slope of line = Rise Run = b a y y = b a (x x ) Rise Run Parametric equation of a line 69

70 1-Tangent planes to the surfaces Parametric equation of a line in two dimensional space passing through the point (x, y ) and parallel to the vector ai + bj is given by x = x + at, y = y + bt Eliminating t from both equation we get x x a = y y b y y = b a (x x) Parametric vector form: r(t) = (x +at) i +(y +bt) j, Equation of line in three dimensional Parametric equation of a line in three dimensional space passing through the point (x, y, z ) and parallel to the vector ai + bj + ck is given by x = x + at, y = y + bt, z = z + ct Eliminating t from these equations we get x x = y y = z z a b c EXAMPLE Parametric equations for the straight line through the point A (, 4, 3) and parallel to the vector v = 4i + j 7k. x =, y = 4, z = 3 and a = 4, b =, c = - 7. The required parametric equations of the straight line are x = + 4t, y = 4 + t, z = 3 7t Different form of equation of curve Curves in the plane are defined in different ways Explicit form: y = f(x) 7

71 1-Tangent planes to the surfaces Example Implicit form: F(x, y) = Example ] Parametric form: Example y = 9 - x, -3 x 3. x + y = 9, - 3 x 3, y 3 Parametric vector form: Equation of a plane x= f(t) and y = g(t) x = 3cos θ, y = 3sinθ, θ x = 3 cos θ, y = 3 sinθ x + y = 9 cos θ + 9sin θ = 9(Cos θ + Sin θ) x + y = 9 A plane can be completely determined if we know its one point and direction of perpendicular (normal) to it. r(t) = f(t) i +g(t) j, a t b. : r ( t) = 3 cos θ i + 3 sinθj, < θ <. Let a plane passing through the point P (x, y, z ) and the direction of normal to it is along the vector n = ai + bj + ck Let P (x, y, z) be any point on the plane then the line lies on it so that n PP ( means perpendicular to ) P P = (x x ) i + ( y y ) j + (z z ) k n. P P = 71

72 1-Tangent planes to the surfaces a (x x) + b (y y) + c (z z) = is the required equation of the plane Here we use the theorem,let a and b be two vectors, if a and b are perpendicular then a.b= so n and PP are perpendicular vector so n.pp= REMARKS Point normal form of equation of plane is a (x x ) + b (y y ) + c (z z ) = We can write this equation as ax + by + cz ax by cz = ax + by + cz + d = where d = ax by cz Which is the equation of plane EXAMPLE An equation of the plane passing through the point (3, - 1, 7) and perpendicular to the vector n = 4i + j - 5k. A point-normal form of the equation is 4(x 3) + (y + 1) 5 (z 7) = 4x + y 5z + 5 = Which is the same form of the equation of plane ax + by + cz + d = The general equation of straight line is ax + by + c = Let (x1, y 1 ) and (x, y ) be two points on this line then ax1 + by1 + c = ax + by + c = Subtracting above equation a(x x1 ) + b (y y 1 ) = 7

73 1-Tangent planes to the surfaces v = (x x 1 )i + (y y 1 )j is a vector in the direction of line φ(x, y) = ax + by φ x = a, φ y = b φ = ai + bj = n φ. r = Then n and v are perpendicular The general equation of plane is ax + by + cz + d = For any two points (x 1, y 1, z 1 ) and (x, y, z ) lying on this plane we have ax 1 + by 1 + cz 1 + d = (1) a x + by + cz + d = () Subtracting equation (1) from () have a (x x 1 )+b (y y1 ) + c (z z 1 ) = (a i + b j + c k ). [ (X -X 1 ) i+(y -Y 1 ) j +(Z -Z 1 ) k ] Here we use the definition of dot product of two vectors. φ = ax + by + cz φx = a, φy = b, φz = c φ = ai + bj + ck Where v = (x x 1 )i +(y y 1 )j + (z z 1 )k φ is always normal to the plane. Gradients and Tangents to Surfaces f (x, y) = c z = f(x,y), z = c If a differ entiable function f(x,y) has a constant value c along a smooth curve having parametric equation x = g(t), y = h(t), r = g(t) i + h (t)j d ifferentiating both sides of this equation with respect to t leads to the equation d dt f (g(t), h(t)) = d dt ( c ) 73

74 1-Tangent planes to the surfaces f dg x dt + f dh y dt = f x i + f y j. dg dt i + dh dt j = d r f. dt = Chain Rule f is normal to the tangent vector d r/ dt, so it is normal to the curve through (x, y ). Tangent Plane and Normal Line Consider all the curves through the point P(x, y, z) on a surface f(x, y, z) =. The plane containing all the tangents to these curves at the point P(x, y, z) is called the tangent plane to the surface at the point P. The straight lines perpendicular to all these tangent lines at P is called the normal line to the surface at P if fx, fy, fz are all continuous at P and not all of them are zero, then gradient f (i.e fxi + fyj + fzk) at P gives the direction of this normal vector to the surface at P. Tangent plane Let P (x, y, z ) be any point on the Surface f(x,y,z) =. If f(x,y,z) is differentiable at po(x,y,z) then the tangents plane at the point P (x,y,z ) has the equation EXAMPLE 9x + 4y z = 36 P (,3,6). f(x,y,z) = 9x + 4y z 36 f x = 18x, f y = 8 y, f z = z Equations f of Tangent Plane to the surface x (P) = 36, f y (P) = 4, f z (P) = -1 74

75 1-Tangent planes to the surfaces through P is 36(x ) + 4(y 3) 1(z 6) = 3x + y z 6 = EXAMPLE z = x cos y ye x (,,). cos y ye x z = f (x,y,z) = cos y ye x z f x (,, ) = (cos y ye x )(,) = 1.1 = 1 f y (,, ) = ( x sin y e x )(,) = 1 = 1. f z (,, ) = - 1 The tangent plane is f x (,,)(x )+f y (,,)(y ) + f z (,,)(z )= 1 (x ) 1 (y ) 1 (z ) =, x y z =. 75

76 13-Orthogonal surfaces Lecture No -13 Orthogonal Surface In this Lecture we will study the following topics Normal line Orthogonal Surface Total differential for function of one variable Total differential for function of two variables Normal line Let P (x,y,z) be any point on the surface f(x,y,z)= If f(x,y,z) is differentiable at P o (x,y,z ) then the normal line at the point P(x,y o,z ) has the equation Here fx means that the function f(x,y,z) is partially differentiable with respect to x And fx(p) means that the function f(x,y,z) is partially differentiable with respect to x at the point P(x,y,z) fy means that the function f(x,y,z) is partially differentiable with respect to y And fy(p) means that the function f(x,y,z) is partially differentiable with respect to y at the point P(x,y,z) Similarly fz means that the function f(x,y,z) is partially differentiable with respect to z And fz(p) means that the function f(x,y,z) is partially differentiable with respect to z at the point P(x,y,z) EXAMPLE x = x +f x (P )t, y = y+fy(p )t, z = z+fz(p )t Find the Equation of the tangent plane and normal of the surface f(x,y,z)= x+y+z-4 at the point P(1,-,3) f(x,y,z) = x +y +z 14 P (1,, 3). f x = x, f y = y, fz = z f x (p) =, f y (P ) = 4, fz (P) = 6 Equation of the tangent p lane to the surface at P is 76

77 13-Orthogonal surfaces (x 1) 4 (y + ) + 6 (z 3) = x y + 3z 14 = Equations of the normal line of the surface through P are x 1 = y + 4 = z 3 6 x 1 = y + = z 3 EXAMPLE 1 3 Find the equation of the tangent plane and normal plane 4x y + 3z = 1 P (, 3,1) f(x,y,z) = 4x y + 3z 1 f x = 8x, f y = y, f z = 6z f x (P) = 16, f y (P) = 6, f z (P) = 6 Equations of Tangent Plane to the surface through P is 16(x ) + 6 (y+3) + 6 (z 1) = 8x + 3y + 3z = 1 Equations of the normal line to the surface through P are x = y + 3 = z x = y + 3 = z Example z = 1 x 7 y - f(x,y,z) = 1 x 7 y - z f x = 7 x 6.y -, f y = - x 7.y -3, f z = -1 f x (, 4, 4) = 7 ()6 (4) - = 14 f y (, 4, 4) = () 7 (4) -3 = f z (, 4, 4) =-1 Equation of Tangent at (, 4, 4) is given by f x (,4,4)(x )+ f y (,4,4)(y 4)+ f z (,4,4)(z 4) = 14 (x ) + ( ) (y 4) (z 4) = 14x y z 16 = The normal line has equation s x = +f x (,4,4) t, y = 4 +f y (,4,4) t, z = 4+f z (,4,4)t x = + 14t, y = 4 t, z = 4 t 77

78 13-Orthogonal surfaces ORTHOGONAL SURFACES Two surfaces are said to be orthogonal at a point of their intersection if their normals at that point are orthogonal. They are Said to intersect orthogonally if they are orthogonal at every point common to them. CONDITION FOR ORTHOGONAL SURFACES Let (x, y, z) be any point of intersection of f (x, y, z) = ---- (1) and g (x, y, z) = -----() Direction ratios of a line normal to (1) are f x, f y, f z Similarly, direction rations of a line normal to () are g x, g y, g z l The two normal lines are orthogonal if and only if f x g x + f y g y + fzgz = EXAMPLE Show that given two surfaces are orthogonal or not f(x,y,z) = x + y + z 16 (1) g(x,y,z) = x + y 638 () Adding (1) and () x + y = 63 4, z = 1 4 (3) f x = x, f y = y f z = 1 g x = x, g y = y g z = 63 f x g x + f y g y + f z g z = 4 (x + y ) 63 using (3) f x g x + f y g y + f z g z = = Since they satisfied the condition of orthogonality so they are orthogonal. Differentials of a functions For a function y = f(x) dy = f / (x) d x is called the differential of functions f(x) d x the differential of x is the same as the actual change in x i.e. dx = x where as dy the differential of y is the approximate change in the value of the functions w hich is different from the actual change y in the value of the functions. 78

79 13-Orthogonal surfaces Distinction between the increment n Δy and the differential dy Approximation to the curve If f is differentiable at x, then the tangent line to the curve y = f(x) at x is a reasonably good approximation near x. Since the tangent line passes to the curve y = f(x) for value of x through the point (x, f(x)) and has its equation is y f / (x) = f(x)(x x) or y = f(x ) + f / (x) (x x) slope f / (x), the point-slope form of EXAMPLE f(x) = x x = 4 and dx =Δx = 3 y = 3 Δ y = x + Δx x = If y = x, then dy dx = 1 x so dy = 1 x dx = 1 x (3) = 3 4 =.75 79

80 13-Orthogonal surfaces EXAMPLE Using differentials approximation for the value of cos 61. Let y = cosx and x = 6 then dx = 61 6 = 1 Δy dy = sinx dx = sin6 (1 ) = Now y = cos x y+δ y = cos (x+δx) = cos (x+dx) = cos (6 +1 )=cos61 cos61 = y+ Δ y = cosx + Δy 3 cos cos = = cos EXAMPLE A box with a square base has its hei ght twice is width. If the width of the box is 8.5 inches with a possible error of ±.3 inches. Let x and h be the width and the height of the box respectively, then its volume V is given by V = x h Since h = x, so (1) take the form V = x 3 dv = 6x dx Since x = 8.5, dx = ±.3, so putting these values in (), we have dv = 6 (8.5) (±.3) = ± 13.5 This shows that the possible error in the volume of the box is ±13.5. TOTAL DIFFERENTIAL If we move from (x, y ) to a point (x + dx, y + dy) nearby, the resulting differential in f is df = fx (x, y ) dx + fy (x, y) dy This change in the linearization of 8

81 13-Orthogonal surfaces EXACT CHANGE Area = xy x = 1, y = 8 Area = 8 x = 1.3 y = 8. Area = Exact Change in area = f is called the total differential of f. =.446 EXAMPLE A rectangular plate expands in such a way that its length changes from 1 to 1.3 and its breadth changes from 8 to 8.. Which is exact Change Let x and y the length and breadth of the rectangle respectively, th en its area is A = xy da = A x dx + A y dy = ydx + xdy By the given conditions x = 1, dx =.3, y = 8, dy =.. da = 8(.3) + 1(.) =.44 EXAMPLE The volume of a rectangular parallelepiped is given by the formula V = xyz. If this solid is compressed from above so that z is decreased by % while x and y each is increased by.75% approximately V = xyz dv = V xd x + Vydy + Vzdx dv = yzdx + xzdy + zydz (1) dx = x, dy = 1 1 y, dz= 1 z Putting these values in (1), we have dv =.75 1 xyz xyz 1 xyz =.5.5 xyz = 1 1 V This shows that there is.5 % decrease in the volume.. 81

82 13-Orthogonal surfaces EXAMPLE A formula for the area Δ of a triangle is Δ = 1 ab sin C. Approximately what error is made in computing Δ if a is taken to be 9.1 instead of 9, b is taken to be 4.8 instead of 4 and C is taken to be 3 3 / instead of 3. By the given conditions a = 9, b = 4, C = 3, da = =.1, db = =.8 dc = = 3 6 = radians Putting these values in (1), we have Δ = 1 ab sin C dδ = a 1 ab sin C da + b 1 ab sin C db + C 1 ab sin C dc dδ = 1 b sin Cda + 1 a sin C db + 1 ab cos CdC dδ= 1 4 sin 3 (.1)+1 9 sin 3 (.8) cos 3 36 dδ= 1 (.1)+1 1 (.8) =.93 % change in area =.93 1 Δ =.93 1 = 3.5% 9 8

83 14-Extrema of functions of two variables Lecture No -14 Extrema of Functions of Two Variables In this lecture we shall fined the techniques for finding the highest and lowest points on the graph of a function or, equivalently, the largest and smallest values of the function. The graph of many functions form hills and valleys. The tops of the hills are relative maxima and the bottom of the valleys are called relative minima. Just as the top of a hill on the earth s terrain need not be the highest point on the earth, so a relative maximum need not be the highest point on the entire graph. Absolute maximum A function f of two variables on a subset of R is said to have an D absolute (global) maximum value on D if there is some point D such that value of f on D In such a case ( x, y ) of f ( x, y ) > f ( x, y ) for all ( x, y ) D f ( x, y ) is the absolute maximum Relative extremum and absolute extremum If f has a relative maximum or a relative minimum at (x, y ), then we say that f has a relative extremum at (x,y ), and if f has an absolute maximum or absolute minimum at (x,y ), then we say that f has an absolute extremum at (x,y ). Absolute minimum A function f of two variables on a subset D of R minimum value on D if there is some point ( x, y f ( x, y ) f ( x, y) for all ( x, y ) D. In such a case f ( x, y ) is the absolute minimum value of f on D. R elative (local) maximum The function f is said to have a relative (local ) maximum at some point (x,y) of its domain D if there exists an open disc K centered at (x,y) and of radius r K ={( x, y ) R : (x x) + ( y y) < r } With such that K D f(x, y ) f (x, y) for all (x, y ) is said to have an ) of D such that absolute (global) Relative (local) minimum 83

84 14-Extrema of functions of two variables The function f is said to have a relative (local) minimum at some point ( x, y ) of D if there exists an open disc K centred at ( x, y ) and of radius r with K D such that f ( x, y ) f (x, y) for all (x, y) K. Extreme Value Theorem If f (x, y) is continuous on a closed and bounded set R, then f has both an absolute maximum and on absolute minimum on R. Remarks If any of the conditions the Extreme Value Theorem fail to hold, then there is no guarantee that an absolute maximum or absolute minimum exists on the region R. Thus, a discontinuous function on a closed and bounded set need not have any absolute extrema, and a continuous function on a set that is not closed and bounded also need not have any absolute extrema. Extreme values or extrema of f The maximum and minimum values of f are referred to as extreme values of extrema of f.let a function f of two variables be defined on an open disc K x, y x x) y y) r = {( ): ( + ( < }. Suppose f x ( x, y ) and f ( x y, y ) both exist on K If f has relative extrema at (x,y),then f x (x, y ) = = fy(x, y). 84

85 14-Extrema of functions of two variables Saddle Point A differentiable function f(x, y) has a saddle has a saddle point (a, b) if in every open disk centered at (a, b) there are domain points (x, y) where f (x, y) > f (a, b) and domain points (x, y) where f (x, y) < f (a, b). The corresponding point (a, b, f (a, b)) on the surface z = f (x, y) is called a saddle point of the surface Remarks Thus, the only points where a function f(x,y) can assume extreme values are critical points and boundary points. As with differentiable functions if a single variable, not every critical point gives rise to o a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variable might have a saddle point. EXAMPLE Fine the critical points of the given function f (x, y) = x 3 + y 3 3axy, a >. f x, f y exist at all points of the domain of f. f x = 3x 3ay, f y = 3y 3ax For critical points f x = f y =. Therefore, x ay = (1) and ax y = () Substituting the value of x from () into (1), we have y 4 a ay = y(y 3 a 3 ) = y=, y= a and so x =, x= a. The critical points are (, ) and (a, a). 85

86 14-Extrema of functions of two variables Overview of lecture # 14 Topic Article # page # Extrema of Functions of Two Variables Absolute maximum Absolute manimum Extreme Value Theorem Exercise set Q#1,3,5,7,9,11,13,15, Book CALCULUS by HOWARD ANTON 86

87 15-Examples Lecture No - 15 Example EXAMPLE f ( x, y) = x + y f x ( x, y ) = f y (x, y) = x x + y y x + y The partial derivatives exist at all points of the domain of in the domain of f. Thus (, ) is a critical point of f Now f x(x, y) = only if x = and f y( x, y ) = only if y = f except at the origin which is The only critical point is (,) and f(,)= Since f (x, y ) for all (x, y), f (, ) = is the absolute minimum value of f. Example z = f(x, y) = x + y (Paraboloid) f x (x, y) = x, f y (x, y) = y when f x (x, y) =, f y (x, y) = we ha ve (, ) as critical point. 87

88 15-Examples EXAMPLE z = g(x, y) = 1 x y (Paraboloid) g x (x, y) = x, g y (x, y) = y when g x (x, y) =, g y (x, y) = we h a ve (, ) as critical point. EXAMPLE z = h(x, y)=y x (Hyperbolic paraboloid) hx (x, y) = x, hy (x, y) = y when hx (x, y) =, h y (x, y) = we ha ve (, ) as critical point. EXAMPLE f(x, y) = x + y x y f x = f y = x + y x + y The point (,) is critical point of f because the partial derivatives do not both exist. It is evident geometrically that fx(,.) does not exist because the trace of the cone in the plane y= has a corner at the origin. The fact that f x (,) does not exist canalso be seen algebraically by noting that fx (,) can be interpreted as thederivative with respect to x of the function f (x, ) = x + = x at x =. 88

89 15-Examples But x is not differentiable at x =, so f x (,) does not exist. Similarly, f y (,) does not exist. The function f has a relative minimum at the critical point (,). The Second Partial Derivative Test Let f be a function of two variables with continuous second order partial derivatives in some circle centered at a critical point (x, y), and let (b) If D > and fxx(x,y) <, then f has a relative maximumat (x,y). (c) If D <, then f has a saddle point at (x,y ). (d) If D =, then no conclusion can be drawn. REMARKS If a function f of two variables has an absolute extremum (either an absolute maximum or an absolute minimum) at an interior point of its domain, then this extremum occurs at a critical point. EXAMPLE D = f xx (x, y ) f yy (x, y ) f xy(x, y ) (a) If D > and f xx (x,y ) >, then f has a relative minimum at (x,y ). f(x,y) = x 4x + xy 1 f x (x, y) = 4x 4 + y, fxx (x, y) = 4 f y (x, y) = xy, fyy (x, y) = x f xy (x, y) = fyx (x, y) = y For critical points, we set the first partial derivatives equal to zero. Then 4x 4 + y = (1) and xy = () we have x = or y = x =, then from (1), y = ±. y =, then from (1), x = 1. Thus the critical points are (1,), (, ), (, ). 89

90 15-Examples We check the nature of each point. f xx(1,) = 4, f yy (1, ) =, f xy (1, ) = D= f xx (1, ).f yy (1, ) - [f xy (1, )] = 8 > and f xx (1, ) is positive. Thus f has a relative minimum at (1, ). f xx(, ) = 4, f yy (, ) =, fxy (, ) = 4 D= f xx (, ).f yy (, ) -[f xy (, )] = 16 <. f xx (, ) = 4, f yy (, ) =, f xy (, ) = 4 D= f xx (, ).f yy (, ) - [f xy (, )] = 16 <. Therefore, f has a saddle point at (,). Therefore, f has a saddle point at (, ). EXAMPLE f(x,y) = e -(x +y+x) f x (x, y)= (x+1)e -(x +y +x), f y(x, y) = ye -(x +y +x) For critical points fx (x,y) =, x + 1 =, x = 1 and fy (x, y) =, y = Hence critical point is ( 1,). f xx (x,y) = [( x ) ]e -(x +y +x) f xx ( 1, ) = -, f yy (x,y) = [4y ]e -(x +y +x) f yy ( 1, ) = - f xy (x,y) = y ( x )e -(x +y +x) f xy ( 1, ) = D = f xx ( 1,) f yy ( 1, ) f xy ( 1, ) = (-e ) (-e ) > This shows that f is maximum at ( 1, ). 9

91 15-Examples EXAMPLE f(x,y) = x 4 + y x y f x (x, y) = 8x 3 x, fy(x, y) = y f xx (x, y) = 4x, f yy(x,y) =, fxy (x, y) = For critical points f x (x, y) =, x (4x 1) =, x =,1/,-1/ f y (x, y) =, y =, y = 1 Solving above equation we have the critical points (,1), 1, 1 1, 1. f xx (,1) =, fyy (, 1) =, f xy (, 1) = D = fx(, 1) fyy (, 1) f xy (, 1) = ( )() = 4 < This shows that (, 1) is a saddle point. 1 1 f xx, 1 = 4, f yy =, 1 = 1 f xy, 1 = 1 1 D = f xx, 1 f yy, 1 f 1 xy, 1 = (4) () = 8 > 1 1 f xx, 1 = 4 >, so f is minimum at, 1. 91

92 15-Examples Example Locate all relative extrema and saddle points of f (x, y) = 4xy x 4 y 4. f x (x, y) = 4y 4x 3, f y (x, y) = 4x 4y 3 For critical points f x (x, y) = 4y 4x 3 = (1) y = x 3 f y (x, y) = 4x 4y 3 = () x = y 3 Solving ( 1 ) and (), we have the critical points (,), (1, 1),( 1, 1). Now f xx (x, y) = 1x, f xx (, ) = f yy (x, y) = 1y, f yy (,) = f xy (x, y) = 4, f xy (, ) = 4 D = f xx (,) f y (,) f xy (,) = () () (4) = - 16 < This shows that (,) is the saddle point. f xx (x, y) = 1x, f xx (1,1) = 1 < f yy (x,y) = 1y, f y (1,1) = 1 f xy (x, y) = 4, fxy (1,1) = 4 D = fxx (1,1) fyy (1,1) f xy (1, 1) = ( 1) ( 1) (4) = 18 > This shows that f has relative maximum at (1,1). f xx (x,y) = 1x, f xx ( 1, 1) = 1 < f y (x, y) = y, f y ( 1, 1) = 1 f xy (x, y) = 4, f xy ( 1, 1) = 4 D=f xx ( 1, 1) f yy ( 1, 1) f xy( 1, 1) = ( 1) ( 1) (4) = 18 > This shows that f has relative maximum ( 1, 1). Over view of lecture # 15 Book Calculus by HOWARD ANTON Topic # Article # Page # Example Graph of f(x,y) The Second Partial Derivative Test Example

93 16-Extreme valued theorem Lecture No -16 Extreme Valued Theorem EXTREME VALUED THEOREM If the function f is continuous on the closed interval [a, b], then f has an absolute maximum value and an absolute minimum value on [a, b] Remarks An absolute extremum of a function on a closed interval must be either a relative extremum or a function value at an end point of the interval. Since a necessary condition for a function to have a relative extremum at a point C is that C be a critical point, we may determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval [a, b] by the following procedure. 1. Find the critical points of f on [a, b] and the function values at these critical.. Find the values of f (a) and f (b). 3. The largest and the smallest of the above calculated values are the absolute maximum value and the absolute minimum value respectively Example Find the absolute extrema of f(x)= x 3 + x -x+1 on [-,1/] Since f is continuous on [-,1/], the extreme value theorem is applicable. For this f / (x) =3 x +x-1 This shows that f(x) exists for all real numbers, and so the only critical numbers of f will be the values of x for which f (x)=.. Setting f / (x) =, we have (3x 1) (x + 1) = from which we obtain x = 1 and x = 1 3 The critical points of f are 1 and 1 3, and each of these points is in the given closed interval 1 (-, ) We find the function values at the critical points and at the end points of the interval, which are given below. f( ) = 1, f ( 1) =, 1 f 3 = 7, f 1 = 7 8 The ab solute maximum value of f on (-, 1 ) is therefore, which occurs at 1, and the absolute min. value of f on (-, 1 ) is 1, which occurs at the left end point. Find the absolute extrema of 93

94 16-Extreme valued theorem f (x) = (x ) /3 on [1, 5]. Since f is continuous on [1. 5], the extreme- value t theorem is applicable. Differentiating f with respect to x, we get / f (x) = 3 ( x ) 1/3 There is no value of x for which f (x) =. we conclude that is a critical point of f, so that the absolute extrema occur either at or at one of the end points of the interval. The function values at these points are given below. 3 9 / / However, since f (x) does not exist at, f(1) = 1, f() =, f(5)= From these values we conclude that the absolute minimum value of f on [1,5] is, occurring at, and the absolute maximum value of f on [1, 5] is 3 9,occurring at 5. Find the absolute extrema of h(x) = x /3 on [, 3]. h / (x) = 3 x - 1/3 = 3x 1/3 h / (x) has no zeros but is undefined at x =. The values of h at this one critical point and at the endpoints x = and x = 3 are h() = h ( ) = ( ) /3 = 4 1/3 h(3) = (3) /3 = 9 1/3. 1/3 The absolute maximum value is 9 assumed at x = 3; the absolute minimum is, assumed at x =. How to Find the Absolute Extrema of a Continuous Function f of Two Variables on a Closed and Bounded Region R. Step 1. Find the critical points of f that lie in the interior of R. Setp. Find all boundary points at which the absolute extrema can occur, Step 3. Evaluate f(x,y) at the points obtained in the previous steps. The largest of these values is the absolute maximum and the smallest the absolute minimum. 94

95 16-Extreme valued theorem Find the absolute maximum and minimum value of f(x,y) = + x +y-x -y On the triangular plate in the first quadrant bounded by the lines x=,y=,y=9-x Since f is a differentiable, the only places where f can assume these values are points inside the triangle having vertices at O(,), A(9,)and B(,9) where f x = f y = and points of boundary. B (, 9) x =.(1,1).(9/,9/) y = 9 - x O For interior points: y = A( 9,) We have f x =-x= and f y =-y= yielding the single point (1,1) For boundary points we take take the triangle one side at time : 1. On the segment OA, y= U(x) = f(x, )=+x-x may be regarded as function of x defined on the closed interval x 9 Its extreme values may occur at the endpoints x= and x=9 which corresponds to points (, ) and (9, ) and U(x) has critical point where U / (x) = -x= Then x=1 On the segment OB, x= and V(y)=f(,y)= +y-y Using symmetry of function f, possible points are (, ),(,9) and (,1) 3. T he interior points of AB. With y = 9 - x, we have f (x, y) = +x+(9 - x) x (9- x) W(x) = f(x, 9 - x) = x x Setting w (x)= 18-4x =, x = 9/. At this value of x, y = 9 9/ Therefore we have ( 9, 9 ) as a critical point. (x, y) 9 (,) (9,) (1, ), 9 f(x,y)

96 16-Extreme valued theorem (x, y) (, 9) (,1) (1,1) f(x,y) The absolute maximum is 4 which f assumes at the point (1,1) The absolute minimum is -61 which f assumes at the points (, 9) and (9,) EXAMPLE Find the absolute maximum and the absolute minimum values of f(x,y)=3xy-6x-3y+7 on the closed triangular region r with the vertices (,), (3,) and (,5). f(x, y) = 3xy 6x 3y + 7 f x (x, y) = 3y 6, f y (x, y) = 3x 3 For critical points f x (x, y) = 3y 6 = y = f y (x, y) = 3x 3 = x = 1 Thus, (1, ) is the only critical point in the interior of R. Next, we want to determine the location of the points on the boundary of R at which the absolute extrema might occur. The boundary extrema might occur. The boundary each of which we shall treat separately. i)the line segment between (, ) and (3, ): On this line segment we have y= so (1) simplifies to a function of the single variable x, u (x)=f(x, ) = 6x + 7, < x < 3 This function has no critical points because u / (6)=-6 is non zero for all x. Thus, the extreme values of u(x) occur at the endpoints x = and x=3, which corresponds to the points (, ) and (3,) on R ii) The line segment between the (,) and (,5) On this line segment we have x=,so single variable y, v(y) = f (, y) = 3y + 7, < y < 5 This function has no critical points because v / (y)=-3 is non zero for all y.thus,the extreme values of v(y) occur at the endpoints y = and y=5 which correspond to the point (,) and (,5) or R 96

97 16-Extreme valued theorem iii) The line segment between (3,) and (,5) In the XY- plan, an equation for the line segment 5 y = 3 x + 5, < x < 3 so (1) simplifies to a function of the single variable x, 5 w(x) = f (x, 3 x + 5) = 5x + 14x 8, < x < 3 w (x) = 1 x + 14 w (x) = 1x + 14 = x = 7 5 This shows that x =7/5 is the only critical point of w. Thus, the extreme values of w occur either at the critical point x=7/5 or at the endpoints x= and x=3.the endpoints correspond to the points (, 5) and (3, ) of R, and from (6) the critical point corresponds to [7/5,8/3] (x, y) (, ) (3, ) (, 5) 7 5, 8 3 (1, ) f(x,y) Finally, table list the values of f(x,y) at the interior critical point and at the points on the boundary where an absolute extremum can occur. From the table we conclude that the absolute maximum value of f is f(,)=7 and the absolute minimum values is f(3,)=-11. OVER VIEW: Maxima and Minima of functions of two variables. Page # 833 Exercise: 16.9 Q #6,7,8,9. 97

98 18-Revision of integration Lecture No -17 Examples EXAMPLE Find the absolute maximum and minimum values of f(x,y)=xy-x-3y on the closed triangular region R with vertices (, ), (, 4), and (5, ). f (x,y) = xy x 3y (1) f x (x, y) = y 1, f y (x, y) = x 3 For critical points f x (x, y) =, y 1 = y = 1 () f y (x, y) =, 3x 3 = x = 3 (3) Thus, (3, 1) is the only critical point in the interior of R. Next, we want to determine the location of the points on the boundary of R at which the absolute extrema might occur. The boundary of R consists of three line segments, each of which we shall treat separately. (i) The line segment between (, ) and (5,) On this line segment we have y =, so (1) simplifies to a function of the single variable x, u (x) = f (x, ) = - x, <x < 5 (4) The function has no critical points because the u / (x)=-1 is non zero for all x.thus,the exteme values of u(x) occure at the endpoints x= and x=5, which corresponds to the points (,) and (5,) of R. ii) The line segment between (,) and (,4) On this line segment we have x =, so (1) simplifies to a function of the single variable y, v(y) = f (, y) = 3y, < y < 4. (5) This function has no critical points because v / (y)= -3 is nonzero for all y. Thus, the extreme values of v(y) occur at the endpoints y = and y=4,which correspond to the point (,) and (,4) or R. iii) The line segment between (5,) and (,4) In the xy-plan, an equation is y 4 = x+ 4, < x < 5 (6) 5 so (1) simplifies to a function of the single variable x, 4 wx ( ) = f( x, x+ 4) = x( x+ 4) x 3( x+ 4) = x + x

99 18-Revision of integration 8 7 w ( x) = x by w ( x) =, we get x = 8 7 This shows that x = is the only critical point of w. Thus, the extreme values of w 8 7 occur either at the critical point x = or at the endpoints x = and x = 5. The endpoints 8 correspond to the points (, 4) and (5, ) of R, and from (6) the critical point corresponds 7 13 to, 8 1 ( x, y ) (, ) (5, ) (, 4) (7/8, 13/1) (3, 1) f ( x, y ) /8-3 Finally, from the table below, we conclude that the absolute maximum value of f is f (,) = and the absolute minimum value is f (, 4) =-1 Example Find three positive numbers whose sum is 48 and such that their product is as large as possible Let x,y and z be the required numbers, then we have to maximize the product f(x,y)=xy(48-x-y) Since f x =48y-xy-y, f y =48x-xy-x solving f x =, f y = we get x=16, y=16, z=16 Since x+y+z=48 f xx (x,y) = - y, f xx (16, 16 ) = - 3 < f xy (x, y ) = 48 - x - y, f xy (16, 16 ) = - 16 f yy (x, y ) = - x, f yy (16, 16 ) = - 3 D=f xx (16,16 )f y7y (16,16) f xy (16,16) = ( 3) ( 3) (16) = 768 > For x = 16, y = 16 we have z = 16 since x + y + z = 48 Thus, the required numbers are 16, 16, 16. Example Find three positive numbers whose sum is 7 and such that the sum of their squares is as small as possible 99

100 18-Revision of integration Let x, y, z be the required numbers, then we have to f(x,y) = x + y + z = x + y + (7 x y) Since x+y+z = 7 f x = 4x+y 54, f y = x+4y 54, f xx = 4, f yy = 4, f xy = Solving f x =, f y = W e get x = 9, y = 9, z = 9 Since x + y + z = 7 D = f xx (9, 9) f yy (9, 9) [f xy (9, 9)] = (4) (4) =1 > This shows that f is minimum x = 9, y = 9, z = 9, so the required numbers are 9, 9, 9. Example Find the dimensions of the rectangular box of maximum volume that can be inscribed in a sphere of radius 4. Solution: The volume of the parallelepiped with dimensions x, y, z is V = xyz Since the box is inscribed in the sphere of radius 4, so equation of sphere is x + y + z = 4 from this equation we can write z = 16 x y and putting this value of z in above equation we get V = xy 16 x y.now we want to find out the maximum value of this volume, for this we will calculate the extreme values of the function V. For extreme values we will find out the critical points and for critical points we will solve the equations V x = and V y =.Now we have xy( x) Vx = y 16 x y + 16 x y x y + 16 x y + 16 Vx = y Now V x = y = 16 x y 16 x y x y + 16 = x + y = 16...( a) Similarly we have xy( y) Vy = x 16 x y + 16 x y x y + 16 x y + 16 Vy = x Now V y = x = 16 x y 16 x y + 16 = + = 16...( ) x y x y b 1

101 18-Revision of integration 4 4 Solving equations (a) and (b) we get the x = and y = 3 3 xy(x + 3y 48) Now Vxx = (We obtain this by using quotient rule of differentiation) 3 (16 x y ) V xx (, ) = p xy(3x + y 48) Also we have to calculate Vyy = and V 3 yy (, ) = p Also note (16 x y ) that V xy (, ) = Now as we have the formula for the second order partial derivative is f. f ( f ) and putting the values which we calculated above we note xx yy xy that fxx (, ). fyy (, ) ( fxy (, )) =+ f Which shows that the function V has maximum value when x = and y =. So the dimension of the rectangular box are x =, y = and z = Example A closed rectangular box with volume of 16 ft 3 is made from two kinds of materials. The top and bottom are made of material costing Rs. 1 per square foot and the sides from material costing Rs.5 per square foot. Find the dimensions of the box so that the cost of materials is minimized Let x, y, z, and C be the length, width, height, and cost of the box respectively. Then it is clear form that C=1(xy+xy)+5(xz+xz)+5(yz+yz) (1) C=xy+1(x+y)z The volume of the box is given by xyz= () Putting the value of z from () in (1), we have C = xy + 1 (x + y) 16 xy C = xy + 16 y + 16 x C x =y - 16, C y = x x 16 y 11

102 18-Revision of integration For critical points C x = y 16 x = and C y = x 16 y = Solving these equations, we have x =, y =. Thus the critical point is (, ). C xx (x, y) = 3 x 3 C xx (, ) = 3 8 = 4 > C yy (x, y) = 3 y 3 C yy (, ) = 3 8 C xy (x, y) = C xy (, ) = = 4 C xx (,) C yy (,) C xy(,) =(4)(4)-() =1> This shows that S has relative minimum at x = and y =. Putting these values in (), we have z = 4, so when its dimensions are 4. Example Find the dimensions of the rectangular box of maximum volume that can be inscribed in a sphere of radius a. Solution: The volume of the parallelepiped with dimensions x, y, z is V = xyz Since the box is inscribed in the sphere of radius 4, so equation of sphere is x + y + z = 4 from this equation we can write z = a x y and putting this value of z in above equation we get V = xy a x y.now we want to find out the maximum value of this volume, for this we will calculate the extreme values of the function V. For extreme values we will find out the critical points and for critical points we will solve the equations V x = and V y =.Now we have xy( x) Vx = y a x y + a x y x y + a x y + a Vx = y Now V x = y = a x y a x y + = + = Similarly we have x y 16 x y a...( a) 1

103 18-Revision of integration y = + V x a x y xy( y) a x y x y + a x y + a Vy = x Now V y = x = a x y a x y x y + a = x + y = a...( b) a a Solving equations (a) and (b) we get the x = and y = 3 3 xy(x + 3y 3 a ) Now Vxx = (We obtain this by using quotient rule of differentiation) 3 ( a x y ) a a a V xx (, ) = p xy(3x + y 3 a ) a a a Also we have to calculate Vyy = and V 3 yy (, ) = Also note ( a x y ) a a a that V xy (, ) = Now as we have the formula for the second order partial derivative is f. f ( f ) and putting the values which we calculated above we note xx yy xy a a a a a a a that fxx (, ). fyy (, ) ( fxy (, )) =+ f Which shows that the a a function V has maximum value when x = and y =. So the dimension of the 3 3 a a a rectangular box are x =, y = and z = Example: Find the points o the plane x + y + z = 5 in the first octant at which f(x,y,z) = xy z has maximum value. Solution: Since we have f(x,y,z) = xy z and we are given the plane x + y + z = 5 from this equation we can write x = 5 y z. Thus our function f becomes f((5 y z),y,z) = (5 y z )y z Say this function u(y,z) That is u(y,z) = (5 y z )y z Now we have to find out extrema of this function. On simplification we get u(y,z) = 5 y z y 3 z y z 3 u y = 1yz 3y z yz 3 = yz (1 3y z) u z = 1y z y 3 z 3y z = y z (1 y 3z) u y =, u z = y =, z = 1 3y z = 1 y 3z = On solving above equations we get 1 + 5z = z = and 1 3y 4 = y = 13

104 18-Revision of integration u yy = 1z 6yz z 3 u zz = 1y y 3 6y z u yz = yz 6y z 6yz at y =, z = u yy (,) = = 4 < y zz (,) = = 4 u yz (,) = = 16 D = u yy u zz (uyz) = ( 4) ( 4) ( 16) = = 3 > For y = and z = We have x = 5 = 1 Example: Find all points of the plane x+y+z=5 in the first octant at which f(x,y,z)=xy z has a maximum value. f (x,y,z) = xy z = xy (5 x y) Since x+y+z = 5 f x = y (5 3x y)(5 x y), f y = xy(5 x y) (5 x y) Solving f x =, f y =, we get x = 1, y =, z = x + y + z = 5 f xx = y (5 3x y) 3y (5 x 5) f xy =y (5 x y)(5 3x y) y (5 3x y) y (5 x y) f yy =x(5 x y)(5 x y) xy(5 x y) 4xy (5 x y) f xx (1,, ) = 4 < f yy (1,, ) = 16 f xy (1,, ) = 8 f xx f yy (f xy ) = ( 4)( 16) ( 8) = 3 > Hence f has maximum value when x = 1 and y =. Thus the points where the function has maximum value is x = 1,y = and z =. 14

105 18-Revision of integration Lecture No -18 Revision of Integration Example: Consider the following integral 1 ( + ) ( xy + y ) dy = x ydx + y dx xy y dxintegrating we get y y 1 = x + = y( ) + y 3 1 ( xy + y ) dx = y( ) + y Example: Double Integral Consider the following integral 1 ( + ) ( xy + y ) dy = x ydx + y dx 1 ( ) xy y dyintegrating we get x 1 xy + y dx = + 3 y y 1 1 = x + = x( ) Symbolically, the double integral of two variables x and y over the certain region R of the plane is denoted by f ( x, y) dxdy Example:. R Use a double integral to find out the solid bounded above by the plane R= ( x, y): x 1, y Z = 4 x y and below by the rectangle { } Solution: We have to find the region R out the volume V over that is, V = (4 x y) da And the solid is shown in the figure below. R 15

106 18-Revision of integration 1 V = (4 x y) da= (4 x y) dxdy R 1 x= 1 x (4 x y) dxdy = 4x xy dy After putting the upper and lower limits we get x= 7 7 y ydy = y. solid V = (4 x y) da= 5 Example: R again after putting the limits we get the required volume of the Evaluate the double integral 1 1 ( ) xy + y dxdy Solution: First we will integrate the given function with respect to x and our y y integral becomes ( xy + y ) dydx = x + dy 3 and after applying the limits we have, x 1 ( xy + y ) dydx = + dy integrating we get x x ( xy + y ) dydx = + = + = Iterated or Repeated Integral d b The expression f ( x, y) dx dy is called iterated or repeated integral. Often the brackets c a are omitted and this expression is written as d b d b f ( x, y) dxdy = b f ( x, y) dx dy Where we have f ( xydx, ) yields a function of y, c a c a a which is then integrated over the interval c y d. bd b d Similarly f ( x, y) dydx = d f ( x, y) dy dx Where we have f ( xydy, ) yields a function a c a c c of x which is then integrated the interval a x b. Example: Evaluate the integral 1 ( x + 3) dydx. Solution: Here we will first integrate with respect to y and get a function of x then we will integrate that function with respect to x to get the required answer. So 16

107 18-Revision of integration 1 1 ( x + 3) dydx = ( x + 3) y dx and after putting the limits we get, is x ( x + 3) y dx= ( x+ 3) dx= + 3x 1 and the required value of the double integral 1 ( x+ 3) dydx= ( + 3) = 7. Now if we change the order of integration so we get have 1 ( x + 3) dxdy Then we 1 1 x 7 7 ( x+ 3) dxdy = ( + 3) dy = 7 dy = y =.Now you note that the value of the integral remain same if we change the order of integration. Actually we have a stronger result which we sate as a theorem. Theorem: Let R be the rectangle defined by the inequalities a < x < b and c < y < d. If f(x, y) is continuous on this rectangle, then f ( x, y) da = f ( x, y) dxdy = f ( x, y) dxdy d b. b d R c a a c Remark: This powerful theorem enables us to evaluate a double integral over a rectangle by calculating an iterated integral. Moreover the theorem tells us that the order of integration in the iterated integral does not matter. Example: Evaluate the integral ln ln 3 e x+ y dxdy Solution: First we will integrate the function with respect to x. Note that we can write x y e + x y as. ln ln3 ln y x y e e So we have, e e dy = e (3 1) dy Here we use the fact that e and ln are inverse function of each other. So we have ln ln3 ln ln y x y y 3 ln3 e =.Thus we get, e e dy = e dy = e = ( 1) = is the required answer. Example: ln3 ln Evaluate the integral e x+ y dydx(note that in this example we change the order of integration) Solution: First we will integrate the function with respect to y. Note that we can write x y e + x y as. ln 3 ln ln 3 x y x e e So we have, e e dy = e ( 1) dy Here we use the fact that e and ln ln are inverse function of each other. So we have e =.Thus we get, 17

108 18-Revision of integration ln 3 ln ln 3 x y x x ln3 e e dx = e dy = e = (3 1) = is the required answer. Note that in both cases our integral has the same value. Over view: Double integrals Page # Exercise Set 17.1 (page 857): 1,3,5,7,9,11,13,15,17,19 18

109 19-Use of integrals Area as anti-derivatives Lecture No -19 y-axis Use Of Integrals (4,8) 4 x d x = x 4 = (4) = 16 8 Area of triangle =1/ base x altitude = ½ (4) (8) = 16 volume as anti-derivative 3 Volume = 5 d y d x 3 = 5y d x = 15 dx x-axis = 15x = 3 x, y 3, z 5 Volume = x 3 x 5 = 3 The following results are analogous to the result of the definite integrals of a function of single variable. cf(x,y)d xdy R =c R f(x,y)dxdy ( c a constant ) [f(x,y) + g(x,y)] d xdy R = f(x,y) dxdy + g(x, y) dxdy R R [f (x,y) g(x,y)] dxdy R = f(x,y) dxdy g(x, y) dxdy R R Use double integral to find the volume under the surface z = 3x 3 +3x y and the rectangle {(x,y):1 x 3, y }. 3 V olume = 1 ( 3x3 + 3x y) dx dy = 3x 4 3 (3)4 = [ 4 = 4 +x3 y 3 dy (3)3 y y] dy [6 +6 y] dy = 6y +13 y = 17 Use double integral to find the volume of solid in the first octant enclosed by the surface z = x and the planes x=, y =, y = 3 and z = 4 19

110 19-Use of integrals 3 Volume = x d ydx 3 = x y d x = [3x ] dx = x 3 = 8 R f(x,y)da if f(x,y) on R If f(x,y) is nonnegative on a region R, then subdividing R into two regions R 1 and R has the effect to subdividing the solid between R and z= f(x,y) into two solids, the sum of whose volumes is the volume of the entire solid f (x, y) da = f(x, y) da+ f (x, y)da R R 1 R The volume of the solid S can also be obtained using cross sections perpendicular to the y - axis. d Vol (S) = A(y) dy (1) c Where A(y) represents the area of the cross section perpendicular to the y - axis taken at the point y How to compute cross sectional area For each fixed y in the interval c y d, the function f(x,y) is a function of x alone,and A(y)may be viewed as the area under the graph of this function along the interval a<x<b, Thus b A (y) = a f (x, y) dx Substiuting this expression in (1) yields. R f(x,y)da R cg(x,y)da if f(x,y) g(x,y) d b Vol (S) = c f (x, y ) dx dy a d b = Similarly the volume c a f(x, y) dx dy of the side S can also be obtained using sections perpendicular to the x-axis b Vol (S) = A(x) (3) a Where A(x) is the area of the cross section perpendicular to the x-axis taken at the point x. 11

111 19-Use of integrals For each fixed x in the interval a x b the function f(x,y) is a function of alone, so that the area A(x) is given by d A(x) = f(x,y)dy Su bstituting this expression in (3) yields b Vol (S) = a d f (x, y ) dy dx c b = a d f(x,y) dy dx (4) c From eq () and eq (4) we have d R f (x,y) da = c b f(x,y) dx dy a b = a d f(x,y) dy dx c Double integral for non-rectangular region Type I region is bounded the left and right by the vertical lines x=a and x=b and is bounded below and above by continuous curves y=g 1(x) and y=g (x) where g 1(x) g (x) for a x b If R is a type I region on which f(x, y) is continuous, then b g (x) f(x,y)da = f(x,y)dydx (1) R a g 1 (x ) By the method of cross section, the volume of S is also given by 111

112 19-Use of integrals b Vol (S) = a A (x) dx () where A(x) is the area of the cross section at the fixed point x this cross section area extends from g 1(x) to g (x) in the y-direction, g (x) so, A(x) = g 1 (x) f(x, y) dy Substituting this in ( ) we obtain b g Vol (S) = a (x) g 1 (x) f(x, y) dy dx Since the volume ofs is also given by b g f(x, y) da = R a (x) g 1 ( x) f (x, y) dy dx Type II region is bounded below and above by horizontal lines y=c and y=d and is bounded in the left and right by continuous curves x=h 1 (y) and x=h (y) satisfying h 1 (y) h (y) for for c y d.. If R is a type II region on which f (x, y) is continuous, then d h f(x,y)da= R c ( y ) h 1 (y) f(x,y)dx dy d Similarly, the partial definite integral with respect to f(x,y) is evaluated by holding x fixed and integrating with respect to y. c An integral of the form produces a function of x. d c f (x, y) dy 11

113 -Double integral for non-rectangular region Lecture No - Double integral for non-rectangular region Double integral for non-rectangular region Type I region is bounded the left and right by vertical lines x=a and x=b and is bounded below and above by curves y=g 1 (x) and y=g (x) where g 1 (x) g (x) for a x b b g (x) f(x, y) da = R a g 1 (x) f (x, y) dy dx y = g (x) y = g 1(x) a b Type II region is bounded below and above by the horizontal lines y=c and y=d and is bounded on the left and right by the continuous curves x= h 1 (y) and x=h (y) satisfying h 1 (y) h (y) for c y d d h (y) f(x,y) dxdy = R c h 1 (y) f(x,y)dx dy d x =h 1 (y) x=h (y) c Write double integral of the function f(x,y)on the region whose sketch is given ln 8 ln y 1 ln (ln 8) ln8 f(x,y) dx e x f(x,y) y Write double integral of the function f(x,y)on the region whose sketch is given 113

114 -Double integral for non-rectangular region 1 y 1 1 x f(x,y) dx dy f (x, y) dy dx EXAMPLE Draw the region and evaluate an equivalent integral with the order of integration reversed The region of integration is given by the inequalities 4 y y/ 4 = x + x 4 EXAMPLE Evaluate I = and x = x (4x + ) dx dy. y y/ dy x = y + y y y dy = y y3/ y 3 6 y = (4 (4)3 )3/ 8 6 = (8) = 8 4 y y cos x 5 dx dy I = x x = y = 4 (,4) (4x + ) dy dx 4 y cos x5 dy dx x y x and x. The integral is over the region < y < 4, x = y x = y = 4 (,4) y =x x = y O O 114

115 -Double integral for non-rectangular region = x = 4 = y cos x 5 x cos x 5 dx dx 1 1 cos x5. (5x 4 ) dx 1 = 1 sin x 5 = 1 1 sin 3 Evaluate 1/ 1 I = e y dy dx x The integral cannot be evaluated I the given order since e y has no antiderivative. We shall change the order of integration. The region R which integration is performed is given by 1 < x <, y = x and y = 1 This region is also enclosed by x =, x = y and < y < 1 Thus 1 I = = = 1 4 y/ 1 e y dx dy y e y dy 1 e y = 1 4 (e 1) 1 O 1/ y = x Or x = y/ 3 lnx 1 x dy dx Reversing the order of integration 3 e ln3 = ln3 = y x dx dy x ln3 3 dy = 1 [9 e y ] dy e y 1 (3, ln3) 3 y = lnx 115

116 -Double integral for non-rectangular region = 1 9y ey ln3 = 1 9 ln3 e ln3 + e = 1 9 ln = 1 [9 ln3 4 ] = 9 ln3 Over view of Lecture # Book Calsulus By Howard Anton Chapter #!7 Article # 17. Page ( ) Exercise set 17. 1,,3,5,7,35,37,38 116

117 1-Examples Lecture No -1 Examples 1 4 4x 4 = 1 e - y ( y) dy 8 = e - y dy dx Reversing the order of integration 4 y/4 e - y dx dy 4 = x e - y y/4 1 8 Example 4 dy = e - y 4 = 1 8 y 4 e - y dy e - 16 c = e 16 4 y = 4 (1,4) x = y/4 y = 4x Calculate sin x R x da. where R is the triangle in the xy- plane bounded by the x-axis,the line y=x and the line x=1 We integrate first with respect to y and then with respect to x, we find 1 x 1 = sin x x dy dx y sin x ] x y=x dx y= 117

118 1-Examples 1 = sin xdx= cos() = sin xdx= cos() EXAMPLE 1 y/ e x dxdy Since there is no elementary antiderivative of e x, the integral 1 e x dxdy y/ cannot be evaluated by performing the x-integration first. To evaualte this integral, we express is as an equivalent iterated integral with the order if integration reversed. For the inside integration, y is fixed and x varies from he line x = y/ to the line x = 1. For the outside integration, y varies from to, so the given iterated integral is equal to a double integral over the triangular region R. To reverse the order of integration, we treat R as a type I region, which enables us to write the given integral as 1 x e dxdy y By changing the order of integration we get, 1 x e dxdy y 1x x e dydx = = 1 x [ex y] dx y= EXAMPLE = 1 xe x dx = e x 1 = e 1 Use a double integral to find the volume of the solid that is bounded above by the palne Z=4-x-y and below by the rectangle R = {(x,y): < x < 1, < y < } 118

119 1-Examples V= (4 x y) da R 1 = (4 x y) dx dy 1 = 4x x xy dy x= 7 = y dy = 7 y y = 5 EXAMPLE Use a double integral to find the volume of the tetrahedron bounded by the coordinate planes and the plane z=4-4x-y The tetrahedron is bounded above by the plane. z=4-4x-y (1) and below by the triangular region R Thus, the volume is given by V = (4 4x y) da R The region R is bounded by the x-axis, the y-axis, and the line y = x [set z = in (1)], so that treating R as a type I region yields. V = (4 4x y) da R = 1 -x (4 4x y) dy dx = 1 [4y 4xy y ] -x y= dx = 1 (4 8x+4x )dx = 4 3 Find the volume of the solid bounded by the cylinder x+y = 4 and the planes y + z = 4 and z =. The solid is bounded above by the plane z = 4 y and below by the region R within the circle x + y = 4. The volume is given by V = (4 y) da R Treating R as a type I region we obtain 4 - x V = (4 y) dy dx x 4 - x = 1 4y y - = 8 4 x dx - y= - 4-x dx 119

120 1-Examples = 8 x 4-x + 4 x sin-1 EXAMPLE Use double integral to find the volume of the solid that is bounded above by the paraboiled z=9x + y,below by the plane z= and laterally by the planes x =, y =, x = 3, y = 3 Volume = 3 = 18x dx = 6x x = 6 (7) + 8 = 17 (9x + y ) dy dx 3 = 9x y + y3 3 - = 8 sin -1 (1) -sin -1 (-1) = 8[( ) + ( )] =8() = 16 dx 1

121 -Examples Lecture No - Examples EXAMPLE EXAMPLE Use double integral to find the volume of the wedge cut from the cylinder 4x +y =9 by the plane z= and z=y+3 Solution: x y Since we can write 4x + y = 9 as + = 1this is equation of 3 9 ellipse. 9 y 9 y Now the Lower and upper limits for x are x = and x = And upper and lower limits for x are 3 and 3 respectively. So the required volume is given by xy da R R is the region bounded by the Trapezium with the vertices (1, 3) (5, 3) (, 1) (4, 1) S lope of AD = = E quation of line AD y 1 = (x ) y 1 = x + 4 x = y 5 x = y 5 Slope of B C = = Equation of line BC y 1 = (x 4) y 1 = x 8 x = y + 7 x = y ( y+7 )/ ( xy = dx y dy x y+7 )/ y dy = (3y + 3y) dy -( y - 5) / 1 -( y - 5) / 1 = y 3 + 3y 3 1 = (3) 3 3 (3) = y 3 y 3 9 y ( 3) dxdy D(1,3) A(,1) B(4,1) C(5,3) 11

122 -Examples y 9 y = xy + 3x dy 3 9 y 9 y 9 y 9 y = y y + 3 dy 3 = y y y dy = 9 y ( y) dy+ 3 9 y dy =. ( 9 y ) y dy y = [ ] + ydy= 3 sin ( y y + ) = 3 3 EXAMPLE Use double integral to find the volume of solid common to the cylinders x + y =5 and x + z =5 Volume = 8 5 = x 5 5 x y 5-x = 8 (5 x ) dx x dy dx dx 3 = 8 5x x = = 8 = AREA CALCUALTED AS A DOUBLE INTEGRAL = 3 V = 1 da = da (1) R R However, the solid has congruent cross sections taken parallel to the xy-plan so that V = area of base x height=area of R.1 = area of R Combining this with (1) yields the area formula area of R = da () R EXAMPLE Use a double integral to find the area of the region R enclosed between the parabola y = ½ x and the line y=x area of R= da 4 x = dy dx x / R 1

123 -Examples 4 = [y] x 4 dx y=x = / x 1 x dx = x x3 6 = 16 3 Treating R as type II yields. area of R= da dx dy 8 = [ x] x R 8 = y y/ 8 dx y=x / = = y 1 3 y3/ y y dy 4 EXAMPLE Find the area of the region R enclosed by the parabola y=x and the line y=x+ 4 8 = 16 3 y=x+ - 1 y=x dy dx = - 1 [y]x+ dx x = - 1 [x + x ]dx Exercise Set 17. (page 865): 1,,3,5,7,35,37,38 x = x3 + x 3 = 9-1 y = x+ y = x 13

124 3-polar co-ordinate systems Lecture No - 3 Polar Coordinate Systems POLAR COORDIANTE SYSTEMS To form a polar coordinate system in a plane, we pick a fixed point O, called the origin or pole, and using the origin as an endpoint we construct a ray, called the polar axis. After selecting a unit of measurement, we may associate with any point P in the plane a pair of polar coordinates (r, θ), where r is the distance from P to the origin and θ measures the angle from the polar axis to the line segment OP. P(r,θ ) O Origin Polar Axis P(6,45 ) The number r is called the radial distance of P and θ is called a polar angle of P. In the points (6, 45 ), (3, 5 ), (5, 1 ), and (4, 33 ) are plotted in polar coordinate systems. THE POLAR COORDINATES OF A POINT ARE NOT UNIQUE. P(3,5 ) (5,1 ) (4,33 ) For example, the polar coordinates (1, 315 ), (1, 45 ), and (1, 675 ) 1 all represent the same point In general, if a point P has polar co-ordinate (1,315 ) (r, θ ), then for any integer n=,1,,3,. (r,θ +n.36 ) and (r,θ +n.36 1 ) are also polar co-ordinates of p In the case where P is the origin, the line line segment OP reduces to a point, since r =. Because there is no clearly defined polar angle in this 1 case, we will agree that an arbitrary (1,675 ) Polar angle θ may be used. Thus, for every θ may be used. Thus, for every θ,the point (, θ ) is the origin. (1,-45 ) NEGATIVE VALUES OF R When we start graphing curves in polar coordinates, it will be desirable to allow negative values for r. This will require a special definition. For motivation, consider the point P with polar coordinates (3, 5 ) We can reach this point by rotating the polar axis 5 and then moving forward from the origin 3 units along the terminal side of the angle. On the other hand, we can also reach the point P by rotating the polar axis 45 and then moving backward 3 units from the origin along the extension of the terminal side of the angle 14

125 3-polar co-ordinate systems This suggests that the point (3, 5 ) Terminal might also be denoted by ( 3, 45 ), Sides with the minus sign serving to Polar Axis P(3,5 ) indicate that the point is on the extension of the angle s terminal side rather than on the terminal side itself. Polar Axis P(-3,5 ) Since the terminal side of the angle θ + 18 is the extension of the terminal side other angle θ,we shall define. (-r, θ ) and (r, θ + 18 ) to be polar coordinates for the same point. With r=3 and θ =45 in () if follows that (-3, 45 ) and (-3, 5 ) represent the same point. RELATION BETWEEN POLAR AND RECTANGULAR COORDINATES y P(x,y) P(r,θ) r y = r sinθ o θ x= rcos θ x CONVERSION FORMULA FROM POLAR TO CARTESIAN COORDINATES AND VICE VERSA r θ o x P(x, y) =P(r, θ) y x x = r cos θ y = r sin θ x + y = r y/ x =tanθ Example Find the rectangular coordinates of the point P whose polar co-ordinates are (6,135 ) Solution: Substituting the polar coordinates r = 6 and θ =135 P in x = cosθ and y=sinθ yeilds r = x = 6 cos 135 = 6 ( /) = 3 y = 6 sin 135 = 6 ( /) = 3 Thus, the rectangular coordinates of the point P are ( 3, 3 ) 15

126 3-polar co-ordinate systems Example: Find polar coordinates of the point P whose rectangular coordinates are (, 3) Solution: We will find polar coordinates (r, θ ) of P such that r > and θ. r x y = + = + = + = = ( ) ( 3) y 3 1 tanθ = = = 3 θ = tan ( 3) = x 3 From this we have (-, 3 ) lies in the second quadrant of P. All other Polar co-ordinates of P have the form 5 (4, + n) or ( 4, + n) 3 3, Where n is integer LINES IN POLAR COORDIANTES A line perpendicular to the x-axis and passing through the point with xy co-ordinates with (a,) has the equation x=a. To express this equation in polar co-ordinates we substitute x = r cos θ a = rcos θ (1) This result makes sense geometrically since each point P (r, θ) on this line will yield the value a for r cos θ. A line parallel to the x-axis that meets the y-axis it the point with xy-coordinates (, b) has the equation y = b. Substituting y = r sin θ yields. r sin θ = b () as the polar equation of this line. This makes sense geometrically since each point P (r, θ) on this line will yield the value b for r sin θ For Any constant θ, the equation θ = θ (3) is satisfied by the coordinates of all points of the form P (r, θ ), regardless of the value of r. Thus, the equation represents the line through the origin making an angle of θ (radians) with the polar axis. By substitution x = rcosθ and y = rsinθ in the equation Ax +By +C =. We obtain the general polar form of the line, r (A cosθ + B sinθ) + C = 16

127 3-polar co-ordinate systems CIRCLES IN POLAR COORDINATES Let us try to find the polar equation of a circle whose radius is a and whose center has polar coordinates (r, θ ). If we let P(r, θ) be an arbitrary point on the circle, and if we apply the law of cosines to the triangle OCP we obtain r rr cos (θ θ ) + r = a (1) SOME SPECIAL CASES OF EQUATION OF CIRCLE IN POLAR COORDINATES A circle of radius a, centered at the origin, has an especially simple polar equation. If we let r = in (1), we obtain r = a or, since a >, r = a This equation makes sense geometrically since the circle of radius a, centered at the origin, consists of all points P (r, θ) for which r = a, regardless of the value of θ If a circle of radius a has its center on the x-axis and passes through the origin, then the polar coordinates of the center are either (a, ) or (a, ) depending on whether the center is to the right or left of the origin Lecture No -4 Sketching 17

128 5-Double intagrals in polar co-ordinates Draw graph of the curve having the equation r = sin θ By substituting values for θ at increments of (3 ) and calculating r, we can construct 6 The following table: θ (radians) 6 r = sin θ θ (radians) r = sin θ Note that there are 13 pairs listed in Table, but only 6 points plotted in This is because the pairs from θ = on yield duplicates of the preceding points. For example, ( ½, 7/6) and (1/, /6) represent the same point. The points appear to lie on a circle. hat this is indeed the case may be seen by expressing the given equation in terms of x and y. We first multiply the given equation through by r to obtain r = r sin θ which can be rewritten as x + y = y or x + y y = or on completing the square x +.This is a circle of radius 1\ centered at the point (,1/) in the xy-plane. y 1 = 1 4 Sketching of Curves in Polar Coordinates 1.SYMMETRY (i) Symmetry about the Initial Line If the equation of a curve remains unchanged when (r,θ ) is replaced by either (r,-θ ) in its equation,then the curve is symmetric about initial line. (ii) Symmetry about the y-axis If when (r, θ ) is replaced by either (r, θ ) in The equation of a curve and an equivalent equation is obtained,then the curve is symmetric about the line perpendicular to the initial i.e, the y-axis (r,-θ) (r, θ ) (r,-θ) (r, θ ) (ii) Symmetry about the Pole If the equation of a curve remains unchanged (r, θ ) 18

129 5-Double intagrals in polar co-ordinates when either (-r, θ ) or is substituted for (r, θ ) in its equation,then the curve is symmetric about the pole. In such a case,the center of the curve.. Position Of The Pole Relative To The Curve See whether the pole on the curve by putting r= in the equation of the curve and solving for θ. 3. Table Of Values Construct a sufficiently complete table of values. This can be of great help in sketching the graph of a curve. II Position Of The Pole Relative To The Curve. When r =, θ =.Hence the curve passes through the pole. III. Table of Values θ /3 / /3 r=a (1 cosθ) a/ a 3a/ a As θ varies from to, cos θ decreases steadily from 1 to 1, and 1 cos θ increases steadily from to. Thus, as θ varies from to, the value of r = a (1 cos θ) will increase steadily from an initial value of r = to a final value of r = a. On reflecting the curve in about the x-axis, we obtain the curve. θ = r = a CARDIOIDS θ = θ =, r= r = O O A r = a (1 cos θ) r = a (1 + cos θ) (, /) O D C (a, - /) r = a (1 sin θ) O (, - /) r = a (1 + sin θ) CARDIDOIDS AND LIMACONS 19

130 5-Double intagrals in polar co-ordinates r=a+b sin θ, r = a bsin θ r=a+b cos θ, r = a bcos θ The equations of above form produce polar curves called limacons. Because of the heartshaped appearance of the curve in the case a = b, limacons of this type are called cardioids. The position of the limacon relative to the polar axis depends on whether sin θ or cos θ appears in the equation and whether the + or occurs. LEMINSCATE If a >, then equation of the form r = a cosθ, r = a cos θ r = a sin θ, r = a sin θ represent propeller-shaped curves, called lemiscates (from the Greek word lemnicos for a looped ribbon resembling the Fig 8. The lemniscates are centered at the origin, but the position relative to the polar axis depends on the sign preceding the a and whether sin θ or cos θ appears in the equation. le Examp 13

131 5-Double intagrals in polar co-ordinates r = 4 cos θ The equation represents a lemniscate. The graph is symmetric about the x-axis and the y-axis. Therefore, we can obtain each graph by first sketching the portion of the graph in the range < θ < / and then reflecting that portion about the x- and y-axes.the curve passes through the origin when θ = /4, so the line θ = /4 is tangent to the curve at the origin.as θ varies from to /4, the value of cosθ decreases steadily from 1 to, so that rdecreases steadily from to.for θ in the range /4 < θ < /, the quantity cosθ is negative, so there are no real values of r satisfying first equation. Thus, there are no points on the graph for such θ. The entire graph is obtained by reflecting the curve about the x-axis and then reflecting the resulting curve about the y-axis. ROSE CURVES Equations of the form r = a sin nθ and r = a cos n θ represent flower-shaped curves called roses. The rose has n equally spaced petals or loops if n is odd and n equally spaced petals if n is even The orientation of the rose relative to the polar axis depends on the sign of the constant a and whether sinθ or cosθ appears in the equation. SPIRAL A curve that winds around the origin infinitely many times in such a way that r increases (or decreases) steadily as θ increases is called a spiral. The most common example is the spiral of Archimedes, which has an equation of the form. r = aθ (θ > ) or r = aθ (θ < ) In these equations, θ is in radians and a is positive. EXAMPLE Sketch the curve r = θ (θ > ) in polar coordinates. This is an equation of spiral with a = 1; thus, it represents an Archimedean spiral. Since r = when θ =, the origin is on the curve and the polar axis is tangent to the spiral. A reasonably accurate sketch may be obtained by plotting the intersections of the spiral with the x and y axes and noting that r increases steadily as θ increases. The intersections with the x-axis occur when θ =,,, 3,. 131

132 5-Double intagrals in polar co-ordinates at which points r has the values r =,,, 3,.. and the intersections with the y-axis occur when θ =, 3, 5, 7, at which points r has the values r =, 3, 5, 7, Starting from the origin, the Archimedean spirals r = θ (θ > ) loops counterclockwise around the origin. 13

133 5-Double intagrals in polar co-ordinates Lecture No -5 Double integrals in polar co-ordinates Double integrals in which the integrand and the region of integration are expressed in polar coordinates are important for two reasons: First, they arise naturally in many applications, and second, many double integrals in rectangular coordinates are more easily evaluated if they are converted to polar coordinates. The function z = f(r,θ) to be integrated over the region R As shown in the Fig. INTEGRALS IN POLAR COORDIATES When we defined the double integral of a function over a region R in the xy-plane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natural shapes to use because their sides have either constant x-values or constant y-values. In polar coordintes, the natural shape is a polar rectangle whose sides have constant r and θ- values. Suppose that a function f (r, θ) is defined over a region R that is bounded by the ray θ = α and θ = β and by the continuous curves r = r 1 (θ) and r = r (θ). Suppose also that < r 1 (θ) < r (θ) < a for every value of θ between α and β. Then R lies in a fan-shaped region Q defined by the inequalities < r < a and α < θ < β. Then the double integral in polar coordinates is given as θ= β r= r ( θ ) f (r,θ) da = f (, r θ ) drdθ R θ= α r= r1 ( θ) How to find limits of integration from sketch Step 1. Since θ is held fixed for the first integration, draw a radial line from the origin through the region R at a fixed angle θ. This line crosses the boundary of R at most twice. The innermost point of intersection is one the curve r = r 1 (θ) and the outermost point is on the curve r = r (θ). These intersections determine the r-limits of integration. Step. Imagine rotating a ray along the positive x-axis one revolution counterclockwise about the origin. The smallest angle at which this ray intersects the region R is θ = α and the largest angle is θ = β. This yields the θ-limits of the integration. EXAMPLE Find the limits of integration for integrating f (r, θ) over the region R that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1. Solution: Step 1. We sketch the region and label the bounding curves. Step. The r-limits of integration. A typical ray from 133

134 5-Double intagrals in polar co-ordinates The origin enters R where r =1 and leaves where r =1+cos θ. Step 3. The θ-limits of integration. The rays from the origin that intersect R run from θ = to θ =. The integral is = / 1+cosθ 1 EXAMPLE / 1+cosθ -/ 1 f(r, θ) rdr dθ Evalaute sin θ da R f (r, θ) r dr dθ Where R is the region ion the first quadrant that is outside the circle r = and inside the cardioid r = (1+cosθ). Solution: R = sin θ da= / / / 1 r sin θ] (1+cos θ ) (1+cosθ) r= (sin θ) r dr dθ dθ = [(1+cosθ) sin θ sinθ]dθ = 1 3 (1 + cosθ)3 + cosθ / = = 8 3 EXAMPLE Use a double polar integral to find the area enclosed by the three-petaled rose r = sin 3θ. We calculate the area of the petal R in the first quadrant and multiply by three. Solution: /3 sin3 θ A = 3 da = 3 r r dr dθ = /3 sin3θ dr dθ R /3 /3 = 3 sin 3θ dθ = 3 4 (1 cos 6θ) dθ = 3 4 sin 6θ θ 6 /3 = 3 4 θ 3 /3 4 sin 6θ = 1 4 EXAMPLE 134

135 5-Double intagrals in polar co-ordinates Find the area enclosed by the lemniscate r = 4 cosθ. The total area is four times the first-quadrant portion. Solution: A = 4 /4 /4 4 cos θ r dr dθ=a /4 r /4 = 4 cos θ dθ = 4 sin θ] = 4. r= 4cosθ r= CHANGING CARTESIAN INTEGRALS INTO POLAR INTEGRALS dθ The procedure for changing a Cartesian integral f(x, y) dx dy into a polar integral has R two steps. Step 1. Substiute x = r cos θ and y = r sinθ, and replace dx dy by r dr dθ in the Cartesian integral. Step. Supply polar limits of integration for the boundary of R. The Cartesian integral then becomes f(x,y) dx dy= f(rcosθ, rsinθ)r dr dθ R G where G denotes the region of integration in polar coordinates. Notice that dx dy is not replaced by dr dθ but by r dr dθ. EXAMPLE Evalaute the double integral 1 1-x The region of integration is bounded by < y < 1 x and < x < 1 y = 1 x is the circle x + y = 1, r = 1 On changing into the polar coordinates, the given integral is / 1 / r 4 r3 dr dθ = 4 dθ = EXAMPLE Evaluate I = D 1 / (x + y ) dy dx by changing to polar coordinates. / 1 4 dθ = θ 4 = 8 dx dy x + y by changing to polar coordinates, where D is the region in the first quadrant between the circles. x +y =a and x +y =b, <a<b I = / b a / b r dr dθ r = [ln r] a dθ 135

136 5-Double intagrals in polar co-ordinates / = ln b a dθ= θ ln b a EXAMPLE Evaluate the double integral / = ln b a. 1 1 x x y dydx by changing to polar coordinates. ( + ) The region of integration is bounded by < y < y = 1 x is the circle x + y =1, r = 1 On changing into the polar coordinates, the given integral is 1 x and x 1 / 1 / 4 1 / 3 r 1 1 / 1 ( /) /8 rdrdθ = dθ = dθ = θ = =

137 6-Examples Lecture No - 6 EXAMPLES EXAMPLE 4 Evalaute I= 4y-y (x +y )dx dy by changing into polar coordinates. The region of integration is bounded by <x< 4y y and < y < 4 Now x = 4y y is the circle x +y 4y= x + y = 4y.In polar coordinates this takes the form r = 4r sin θ, r = 4 sin θ On changing the integral into polar coordinates, we have / 4sinθ I = / r. r dr dθ = 64 sin 4 θ dθ= = 1 (using Walli s formula) EXAMPLE Evalaute e x +y dy dx.,where R is the semicircular region bounded by the x-axis and R the curve y = 1 x In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate e x +y with respect to either x or y. Substituting x = r cos θ, y = r sinθ, and replacing dy dx by r dr dθ enables us to evaluate the integral as e x +y dy dx = R 1 er r dr dθ = 1 er 1 dθ = 1 (e 1)dθ = EXAMPLE Let R a be the region bounded by the circle x + y = a. Define e-(x +y ) dxdy = lim a R e -(x +y) dx dy To evaluate this improper integral. l= = lim exp { (x + y )} dx dy = lim a a = lim a exp { r }r dr dθ = lim exp { a } = a Da a exp { (x + y )} dx dy 1 (1 exp { 1 a }dθ = lim a (e 1). (1 exp { a } θ EXAMPLE 137

138 6-Examples Prove that e-t 1 dt=. l = - - exp{ (x +y )} dx dy= - exp{ y }[ - exp{ x }dx] dy = [ - exp{ y }dy][ - exp{ x } dx] = [ - exp{ t }dt][ - exp{ t } dt] =lim exp { -a t } dt] = 4lim Hence we have lim a [ exp{-t }dt] THEOREM 4lim a [ a a [ exp{ t } dt ]= lim a a a a -a = /4 exp { t } dt = a [ -a exp{ (x +y )}dx dy4lim Let G be the rectangular box defined by the inequalities a < x < b, c < y < d, k < z < l If f is continuous on the region G, then G a exp { t } dt] a [ b f(x, y, z) dv = a exp{-t }dt] d c l k = f (x, y, z) dz dy dx Moreover, the iterated integral on the right can be replaced with any of the five other iterated integrals that result by altering the order of integration. d = c l k b = l a k b a d f (x, y, z) dy dz dx = c d f (x, y, z) dx dz dy= l c k b a b a l k b f (x, y, z) dy dx dz= a d f (x, y, z) dz dx dy c d l f (x, y, z) dx dy dz c k EXAMPLE Evalaute the triple integral 1xy z 3 dv over the rectangular box G defined by the G inequalities 1 < x <, < y < 3, < z <. We first integrate with respect to z, holding x and y fixed, then with respect to y holding x fixed, and finally with respect to x. 1xy z 3 dv= G -1 = [16xy 3 ] -1 3 y= 3 dx= 1xy z 3 dzdydx = x dx = 16x ] -1 3 [3xy z 4 ] = 648 z= dy dx = xy dy dx 138

139 6-Examples EXAMPLE Evalaute R (x y + z) dx dy dz Region R : 1 = x x x+y (x y + z) dz dy dx= 1 x (x y + z) < x, 1, < y < x, < z < x + y x+y dy dx = (x y+x+y) (x y) dy dx = 1 (3x 3y) dy dx= 3 1 = 3 x4 x6 z dx = 3 x 5 5 x7 1 1 = = 8 35 x x x y y3 3 Example: Evalaute xyzd x dy dz Where S = {(x,y,z):x +y +z <1, x >, y >, z > } S S is the sphere x + y + z = 1.Since x, y, z are al +ve so we have to consider only the +ve octant of the sphere. Now x + y + z = 1. So that z = 1 x y The Projection of the sphere on xy plan is the circle x + y = 1. This circle is covered as y-varies from to 1 x and x varies from to 1. xyz dx dy dz= R 1-x 1 1-x 1-x -y xyz dz dy dx = 1 1-x xy 1-x -y dy dx z 5 1 = xy 1 x y 1 dy dx = 1 x (y x y y 3 ) dy dx 1 = 1 x y x y y4 1-x 1 4 dx= 1 4 x 1 x x (1 x ) (1 x ) dx 1 = 1 8 (x x 3 + x 5 ) dx= 1 8 x x4 + x6 1 6 = = x dx 139

140 7-vector valued functions Lecture No -7 Vector Valued Functions Recall that a function is a rule that assigns to each element in its domain one and only one element in its range. Thus far, we have considered only functions for which the domain and range are sets of real numbers; such functions are called real-valued functions of a real variable or sometimes simply real-valued functions. In this section we shall consider functions for which the domain consists of real numbers and the range consists of vectors in -space or 3-space; such functions are called vector-valued functions of a real variable or more simply vector-valued functions. In -space such functions can be expressed in the form. r (t) = (x (t), y (t)) = x (t) i + y (t) j and in 3-space in the form r (t) = (x (t), y (t), z (t))i+y(t)j+z (t)k where x(t), y(t), and z(t) are real-valued functions of the real variable t. These real-valued functions are called the component functions or components of r. As a matter of notation, we shall denote vector-valued functions with boldface type [f(t), g(t), and r (t) and realvalued functions, as usual, with lightface italic type [f(t), g(t), and r(t)]. EXAMPLE r (t) = (ln t) i + t + j + (cos t)k then the component functions are x(t)=lnt,y(t)= t +,and z (t) = cost The vector that r(t) associates with t = 1 is r(1)=(ln 1) i+ 3j +(cos ) k= 3j k The function r is undefined if t < because ln t is undefined for such t. If the domain of a vector-valued function is not stated explicitly, then it is understood to consist of all real numbers for which every component is defined and yields a real value. This is called the natural domain of the function. Thus the natural domain of a vectorvalued function is the intersection of the natural domains of its components. PARAMETRIC EQUATIONS IN VECTOR FORM Vector-valued functions can be used to express parametric equations in -space or 3-space in a compact form. For example, consider the parametric equations x = x(t), y = y (t) Because two vectors are equivalent if and only if their corresponding components are equal, this pair of equations can be replaced by the single vector equation. x = x(t), y = y (t) x i + y j = x (t)i + y (t) j Similarly, in 3-space the three parametric equations x = x(t), y = y (t), z = z (t) can be replaced by the single vector equation xi + yj + zk = x(t)i + y(t)j + z(t)k if we let r = x i +y j and r(t) = x(t) i + y(t) j in -sapce and let r = xi + yj + zk and r(t) = x(t) i + y(t)j + z(t)k in 3-space, then both () and (4) can be written as r = r(t) which is the vector form of the parametric equations in (1) and (3). Conversely, every vector equation of form (5) can be rewritten as parametric equations by equating components on the two sides. 14

141 7-vector valued functions EXAMPLE Express the given parametric equations as a single vector equation. (a) x = t, y = 3t (b) x = cost, y = sint, z = t (a) Using the two sides of the equations as components of a vector yields. x i + y j = t i + 3t j (b) Proceeding as in part (a) yields xi + yj + zk = (cos t)i + (sin t)j + tk EXAMPLE Find parametric equations that correspond to the vector equation x i + y j + z k = (t 3 + l) i + 3 j + e t k Equating corresponding components yields. x = t 3 + 1, y = 3, z = e t GRAPHS OF VECOR-VALUED FUNCTOINS One method for interpreting a vector-valued function r(t) in -space or 3-space geometrically is to position the vector r = r (t) with its initial point at the origin, and let C be the curve generated by the tip of the vector r as the parameter t varies The vector r, when positioned in this way, is called the radius vector or position vector of C, and C is called the graph of the function r (t) or, equivalently, the graph of the equation r = r (t). The vector equation r = r (t) is equivalent to a set of parametric equations, so C is also called the graph of these parametric equations. EXAMPLE Sketch the graph of the vector-valued function r(t) = (cos t)i + (sin t)j, < t < The graph of r (t) is the graph of the vector equation xi+yj = (cos t)i + (sin t)j, < t < or equivalently, it is the graph of the parametric equations x = cos t, y = sin t ( < t < ) This is a circle of radius 1 that is centered at the origin with the direction of increasing t counterclockwise. The graph and a radius vector are shown in Fig. 141

142 7-vector valued functions EXAMPLE Sketch the graph of the vector-valued function r(t) = (cos t)i+(sin t)j+k, < t < The graph of r(t) is the graph of the vector equation xi+yj+zk=(cost)i+(sint)j+k, <t< or, equivalently, it is the graph o the parametric equations x=cos t, y = sin t, z= ( < t < ) From the last equation, the tip o the radius vecor traces a curve in the plane z =, and from the first two equations and the preceding example, the curve is a circle of radius 1 centered on the z-axis and traced counterclockwise looking down the z-axis. The graph and a radius vector are shown in Fig. EXAMPLE Sketch the graph of the vector-valued function r(t) = (a cos t)i + (a sin t)j + (ct)k where a and c are positive constant. The graph of r(t) is the graph of the parametric equations. x = a cos t, y = a sin t, z = ct As the parameter t increases, the value of z = ct also increases, so the point (x, y, z) moves upward. However, as t increases, the point (x, y, z) also moves in a path directly over the circle. x = a cos t, y = a sin t in the xy-plane. The combination of these upward and circular motions produces a corkscrew-shaped curve that wraps around a right-circular cylinder of radius a centered on the z-axis. This curve is called a circular helix. EXAMPLE Describe the graph of the vector equation r = ( + t)i + 3tj + (5 4t)k The corresponding parametric equations are x = + t, y = 3t, z = 5 4t The graph is the line in 3-space that passes through the point (,, 5) and is parallel to the vector i + 3j 4k. EXAMPLE The graph of the vector-valued function r (t) = t i + t j + t 3 k is called a twisted cubic. Show that this curve lies o the parabolic cylinder y = x, and sketch the graph for t > 14

143 7-vector valued functions The corresponding parametric equations are x = t, y = t, z = t 3 Eliminating the parameter t in the equations for x and y yields y = x, so the curve lies on the parabolic cylinder with this equation. The curve starts at the origin for t = ; as t increases, so do x, y, and z, so the curve is traced in the upward direction, moving away from the origin along the cylinder. GRAPHS OF CONSTANT VECOR-VALUED FUNCTIONS If c is a constant vector in the sense that it does not depend on a parameter, then the graph of r = c is a single point since the radius vector remains fixed with its tip at c. If c = x i + y j (in -space), then the graph is the point (x, y ), and if c = x i + y j + z k (in 3-space), then the graph is the point (x, y, z ). EXAMPLE The graph of the equation r = i + 3j k is the point (, 3, 1) in 3-space. If r(t) is a vector-valued function, then for each value of the parameter t, the expression r(t) is a real-valued function of t because the norm (or length of r(t) is a real number. For example, If r(t) = t i + (t 1) j Then r(t) = t + (t 1) which is a real-valued function of t. EXAMPLE The graph of r (t) = (cos t)i + (sin t)j + k, <t< is a circle of radius 1 centered on the z-axis and lying in the plane z =. This circle lies on the surface of a sphere of radius 5 because for each value of t r(t) = cos t + sin t + t = = 5 which shows that each point on the circle is a distance of 5 units from the origin. 143

144 8-Limits of vector valued functions Lecture No -8 Limits of Vector Valued Functions The limit of a vector-valued functions is defined to be the vector that results by taking the limit of each component. Thus, for a function r(t) = x (t)i + y (t)j in -space we define. lim r(t) = (lim x(t))i + (lim y(t))j t α t α t α and for a function r(t) = x(t)i + y(t)j + z(t)k in 3-space we define. limr(t)=(limx(t))i+(limy(t))j+(limz(t))k t α t α t α t α If the limit of any component does not exist, then we shall agree that the limit of r (t) does not exist. These definitions are also applicable to the one-sided and infinite limits lim t α +, lim t α lim, and lim. It follows from (1) and () that t + t limr(t) = L t α if and only if the components of r(t) approach the components of L as t a. Geometrically, this is equivalent to stating that the length and direction of r (t) approach the length and direction of L as t α CONTINUITY OF VECTOR-VALUED FUNCTIONS The definition of continuity for vector-valued functions is similar to that for real-valued functions. We shall say that r is continuous at t if 1. r (t ) is defined;. limr(t) exists; t t 3. limr(t) = r (t ). t t It can be shown that r is continuous at t if and only if each component of r is continuous. As with real-valued functions, we shall call r continuous everywhere or simply continuous if r is continuous at al real values of t. geometrically, the graph of a continuous vector-valued function is an unbroken curve. DERIVATIVES OF VECOR-VALUED FUNCTIONS The definition of a derivative for vector-valued functions is analogous to the definition for real-valued functions. DEFINITION The derivative r / (t) of a vector-valued function r(t) is defined by r (t+h) r (t) r / (t) = lim h h Provided this limit exists. For computational purposes the following theorem is extremely useful; it states that the derivative of a vector-valued function can be computed by differentiating each components. THEOREM (a) If r(t) = x(t)i + y(t)j is a vector-valued function in -space, and if x(t) and y(t) are differentiable, then r / (t) = x / (t)i + y / (t)j 144

145 8-Limits of vector valued functions (b) If r(t) = x(t)i + y(t)j + z(t)k is a vector-valued function in 3-space, and if x(t), y(t), and z(t) are differentiable, then r / (t) = x / (t)i + y / (t)j + z / (t)k We shall prove part (a). The proof of (b) is similar. Proof (a): r(t + h) r (t) [x(t+h) x(t)] r / (t) = lim h h = lim h h i + lim h [y(t+h) y(t)] h = x / (t)i + y / (t)j As with real-valued functions, there are various notations for the derivative of a vectorvalued function. If r = r (t), then some possibilities are d dr dt [r(t)], dt, r/ (t), and r / EXAMPLE Let r(t) = t i +t 3 j. Find r / (t) and r / (1) r / (t) = d dt [t ] i + d dt [t3 ] j = t i 3t j Substituting t=1 yields r / (1) = i+3j. TAGENT VECTORS AND TANGENT LINES GEOMETRIC INTERPRETATIONS OF THE DERIVATIVE. Suppose that C is the graph of a vector-valued function r(t) and that r / (t) exists and is nonzero for a given value of t. If the vector r / (t) is positioned with its initial point at the terminal point of the radius vector DEFINITION Let P be a point on the graph of a vector-valued function r(t), and let r(t ) be the radius vector from the origin to P If r / (t ) exists and r / (t ), then we call r / (t ) the tangent vector to the graph of r at r(t ) REMARKS Observe that the graph of a vector-valued function can fail to have a tangent vector at a point either because the derivative in (4) does not exist or because the derivative is zero at the point.if a vector-valued function r(t) has a tangent vector r / (t ) at a point onits graph, then the line that is parallel to r / (t ) and passes through the tip of the radius vector r(t ) is called the tangent line of the graph of r(t ) at r(t ) Vector equation of the tangent line is r = r (t ) + t r / (t ) j 145

146 8-Limits of vector valued functions EXAMPLE Find parametric equation of the tangent line to the circular helix x = cost, y = sint, z = 1 at the point where t = /6 To find a vector equation of the tangent line, then we shall equate components to obtain the parametric equations. A vector equation r=r(t) of the helix is xi + yj + zk = (cost)i + (sin t)j + tk Thus, r(t) = (cos t)i + (sin t)j + tk r / (t) = ( sin t)i + (cos t)j + k At the point where t = /6, these vectors are r 6 = 3 i + 1 j + 6 k and r / 6 = 1 i + 3 j + k so from (5) with t = /6 a vector equation of the tangent line is r 6 + t r/ 6 = 3 i + 1 j + 6 k + t 1 i + 3 j + k Simplifying, then equating the resulting components with the corresponding components of r = xi + yj + zk yields the parametric equation. 3 x = 1 t, y = t,z = 6 + t EXAMPLE The graph of r(t) = t i + t 3 j is called a semicubical parabola Find a vector equation of the tangent line to the graph of r(t) at (a) the point (,) (b) the point (1,1) The derivative of r(t) is r / (t) = ti + 3t j (a) The point (, ) on the graph of r corresponds to t =. As this point we have r / () =, so there is no tangent vector at the point and consequently a tangent line does not exist at this point. (b) The point (1, 1) on the graph of r corresponds to t = 1, so from (5) a vector equation of the tangent line at this point is r = r(1) + t r / (1) From the formulas for r (t) and r / (t) with t = 1, this equation becomes r = (i + j) + t (i + 3j) If r is a vector-valued function in -space or 3-space, then we say that r(t) is smoothly parameterized or that r is a smooth function of t if the components of r have continuous derivatives with respect to t and r / (t) for any value of t. Thus, in 3-space r (t) = x(t)i + y (t)j + z(t)k is a smooth function of t if x / (t), y / (t), and z / (t) are continuous and there is no value of t at which al three derivatives are zero. A parametric curve C in -space or 3-space will be called smooth if it is the graph of some smooth vector-valued function. It can be shown that a smooth vector-valued function has a tangent line at every point on its graph. 146

147 8-Limits of vector valued functions PROPERTIES OF DERIVATIVES (Rules of Differentiation). In either -space or 3-space let r(t), r 1 (t), and r (t) be vector-valued functions, f(t) a realvalued function, k a scalar, and c a fixed (constant) vector. Then the following rules of differentiation hold: d dt [c] = d dt [kr(t)] = k d dt [r(t)] d dt [r 1(t) + r (t)]= d dt [r 1(t)]+ d dt [r (t)] d dt [r 1(t) r (t)]= d dt [r 1(t)] d dt [r (t)] d dt [f(t)r(t)] = f(t) d dt [r(t)]+r(t) d dt [f (t)] In addition to the rules listed in the foregoing theorem, we have the following rules for differentiating dot products in -space or 3-space and cross products in 3-space: d dt [r 1(t).r (t)] = r 1. dr dt + dr 1 dt. r (6) d dt [r 1(t) r (t)]=r 1 dr dt +dr 1 dt r (7) REMARK: In (6), the order of the factors in each term on the right does not matter, but in (7) it does. In plane geometry one learns that a tangent line to a circle is perpendicular to the radius at the point of tangency. Consequently, if a point moves along a circular arc in -space, one would expect the radius vector and the tangent vector at any point on the arc to be perpendicular. This is the motivation for the following useful theorem, which is applicable in both -space and 3-space. THEOREM: If r (t) is a vector-valued function in -space or 3-space and r(t) is constant for all t, then r(t). r / (t) = that is, r(t) and r / (t) are orthogonal vectors for all t.it follows from (6) with r 1 (t)=r (t)=r (t) that d dt [r(t).r(t)] = r(t).dr dt + dr dt. r(t) or, equivalently, d dt [ r(t) ] = r(t). dr dt But r(t) is constant, so its derivative is zero. Thus r(t). dr dt That is the r(t) is perpendicular dr dt = that is r(t). dr dt = 147

148 8-Limits of vector valued functions EXAMPLE Just as a tangent line to a circular arc in -space is perpendicular to the radius at the point of tangency, so a tangent line to a curve on the surface of a sphere in 3-space is perpendicular to the radius at the point of tangency. To see that this is so, suppose that the graph of r(t) lies on the surface of the sphere of radius k > centered at the origin.for each value of t we have r(t) =k, r(t). r / (t) = which implies that the radius vector r(t) and the tangent vector r / (t) are perpendicular. This completes the argument because the tangent line, where it exists, is parallel to the tangent vector. INTEGRALS OF VECTOR VALUED FUNCTION (a) If r(t)=x(t)i +y(t) j is a vector-valued function in -space,the we define. r( t) dt = ( x( t) dt) i + ( y( t) dt) j (1 a) b b b r( t) dt = ( x( t) dt) i + ( y( t) dt) j (1 b) a a a (b) If r(t) = x (t) i + y(t) j + z(t)k is a vector - valued function in 3 - space, then we define. r (t)dt=( x(t)dt )i+( y(t) dt) j +( z(t)dt)k b a r(t)dt=( b x(t)dt) i+( b a a y(t)dt )j+( b a z(t)dt)k (a) (b) Let r() t = ti + 3t j ( a) r() t dt ( b) r() t dt r() t dt = (ti + 3 t j) dt = ( tdt) i + ( 3 t dt) j ( t + C) i+ ( t + C ) j= t i+ Ci+ t j+ C j= t i+ t j+ Ci+ C j= t i+ t j+ C WhereC = C i + C j is anarbitrary vector cons tan t of int egration () b r() t dt = (ti + 3 t j) dt = ( tdt) i + ( 3 t dt) j = t i + t j = ( ) i + ( ) j = 4i + 8j 148

149 8-Limits of vector valued functions PROPERTEIS OF INTEGRALS cr(t) dt = c r (t) dt (3) [r 1 (t)+r (t)] dt = r1(t)dt + r (t) dt (4) [r 1 (t) r (t)]dt = r 1 (t) dt r (t) dt (5) These properties also hold for definite integrals of vector-valued functions. In addition, we leave it for the reader to show that if r is a vector-valued function in -space or 3- d space, then dt [ r(t) dt] = r (t) (6) This shows that an indefinite integrals of r(t) is, in fact, the set of antiderivatives of r(t), just as for real-valued functions. If r(t) is any antiderivative or r(t) in the sense that R / (t) = r(t), then r(t) dt = R(t) + C (7) where C is an arbitrary vector constant of integration. Moreover, b a r(t) dt = R(t) ]b a = R(b) R(a). 149

150 9-Change of parameter Lecture No -9 Change of parameter It is possible for different vector-valued functions to have the same graph. For example, the graph of the function r = (3 cos t)i + (3 sin t)j, < t < is the circular of radius 3 centered at the origin with counterclockwise orientation. The parameter t can be interpreted geometrically as the positive angle in radians from the x-axis to the radius vector. For each value of t, let s be the length of the arc subtended by this angle on the circle The parameters s and t are related by t = s/3, < s < 6 if we substitute this in (1), we obtain a vector-valued function of the parameter s, namely r = 3 cos (s/3)i + 3 sin (s/3)j, < s < 6 whose graph is also the circle of radius 3 centered at the origin with counterclockwise orientation.in various problems it is helpful to change the parameter in a vector-valued function by making an appropriate substitution. For example, we changed the parameter above from t to s by substituting t=s/3 in (1). In general, if g is a real-valued function, then substituting t = g(u) in r(t) changes the parameter from t to u. When making such a change of parameter, it is important to ensure that the new vectorvalued function of u is smooth if the original vector-valued function of t is smooth. It can be proved that this will be so if g satisfies the following conditions: 1. g is differentiable.. g / is continuous. 3. g / (u) for any u in the domain of g. 4. The range of g is the domain of r. If g satisfies these conditions, then we call t = g(u) a smooth change of parameter. Henceforth, we shall assume that all changes of parameter are smooth, even if it is not stated explicitly. ARC LENGTH Because derivatives of vector-valued functions are calculated by differentiating components, it is natural do define integrals of vector-functions in terms of components. EXAMPLE If x / (t) and y / (t) are continuous for a < t < b, then the curve given by the parametric equations x = x(t), y = y(t) (a < t < b) (9) has arc length b L = a dx dt + dy dt dt (1) This result generalizes to curves in 3-spaces exactly as one would expect: If x / (t), y / (t), and z / (t) are continuous for a < t < b, then the curve given by the parametric equations x = x(t), y = y(t), z = z(t) (a < t < b) 15

151 9-Change of parameter has arc length b L = a dx dt + dy dt + dz dt dt (1) EXAMPLE Find the arc length of that portion of the circular helix x = cos t, y = sin t, z = t From t = to t = The arc length is L = dx dt + dy dt + dz dt dt = t) + (cos t) + 1 dt = dt = ARC LENTH AS A PARAMETER For many purposes the best parameter to use for representing a curve in -space or 3-space parametrically is the length of arc measured along the curve from some fixed reference point. This can be done as follows: Step 1: Select an arbitrary point on the curve C to serve as a reference point. Step : Starting from the reference point, choose one direction along the curve to be the positive direction and the other to be the negative direction. Step 3: If P is a point on the curve, let s be the signed arc length along C from the reference point to P, where s is positive if P is in the positive direction from the reference point, and s is negative if P is in the negative direction. By this procedure, a unique point P on the curve is determined when a value for s is given. For example, s = determines the point that is units along the curve in the positive direction from the reference point, and s = 3 determines the point that is 3 units along the curve in the negative direction from the reference point. Let us now treat s as a variable. As the value of s changes, the corresponding point P moves along C and the coordinates of P become functions of s. Thus, in -space the coordinates of P are (x(x,), y(s)), and in 3-space they are (x(s), y(s), z(s)). Therefore, in -space the curve C is given by the parametric equations x = x(s), y = y (s) and in 3-space by x = x(s), y = y(s), z = z (s) REMARKS When defining the parameter s, the choice of positive and negative directions is arbitrary. However, it may be that the curve C is already specified in terms of some other parameter t, in which case we shall agree always to take the direction of increasing t as the positive direction for the parameter s. By so doing, s will increase as t increases and vice versa. The following theorem gives a formula for computing an arc-length parameter s when the curve C is expressed in terms of some other parameter t. This result will be used when we want to change the parameterization for C from t to s. 151

152 9-Change of parameter THEOREM (a) Let C be a curve in -space given parametrically by x = x(t), y = y (t) where x / (t) and y / (t) are continuous functions. If an arc-length parameter s is introduced with its reference point at (x(t ), y (t )), then the parameters s and t are related by t s = t dx du + dy du du (13a) (b) Let C be a curve in 3-space given parametrically by x = x(t), y = y(t), z = z(t) where x / (t), y / (t), and z / (t) are continuous functions. If an arc-length parameter s is introduced with its reference point at (x(t ), y(t ), z(t )), then the parameters s and t are related by t s = t dx du + dy du + dz du du (13b) Proof If t > t, then from (1) (with u as the variable of integration rather than t) it follows that t t dx du + dy du du (14) represents the arc length of that portion of the curve C that lies between (x(t ), y(t )) and (x (t), y(t)). If t < t, then (14) is the negative of this arc length. In either case, integral (14) represents the signed arc length s between these points, which proves (13a). It follows from Formulas (13a) and (13b) and the Second Fundamental Theorem of Calculus (Theorem 5.9.3) that in -space. t ds dt = d dt t dx du + dy du du = and in 3-space t dx dt + dy dt ds dt = d dt t dx du + dy du + dz du dx dt = dt + dy dt + dz dt Thus, in -space and 3-space, respectively, ds dx dt = dt + dy dt (15a) ds dt = dx dt + dy dt + dz dt (15b) REMARKS: Formulas (15a) and (15b) reveal two facts worth noting. First, ds/dt does not depend on t ; that is, the value of ds/dt is independent of where the reference point for the parameter s is located. This is to be expected since changing the position of the reference point shifts each value of s by a constant (the arc length between the reference points), and this constant drops out when we differentiate. The second fact to be noted from (15a) and (15b) is that ds/dt > for all t. This is also to be expected since s increases with t by the remark preceding Theorem If the curve C is smooth, then it follows from (15a) and (15b) that ds/dt > for all t. 15

153 9-Change of parameter EXAMPLE x = t + 1, y = 3t (16) using arc length s as a parameter, where the reference point for s is the point (1, ). In formula (13a) we used u as the variable of integration because t was needed as a limit of integration. To apply (13a), we first rewrite the given parametric equations with u in place of t; this gives from which we obtain x = u + 1, y = 3u dx du =, dy du = 3 we see that the reference point (1, ) corresponds to t = t = t t s = t dx du + dy du du = 13 du = 13u t ] u=t = 13t u= 1 Therefore, t = 13 s Substituting this expression in the given parametric equations yields. x = 1 13 s + 1 = 13 s + 1 y = s = 3 13 s EXAMPLE Find parametric equations for the circle x = a cos t, y = a sin t ( < t < ) using arc length s as a parameter, with the reference point for s being (a, ), where a >. We first replace t by u in the given equations so that x = a cos u, y = a sin u And dx du = a sin u, dy du = a cos u Since the reference point (a, ) corresponds to t =, we obtain t t s = t dx du + dy du du = ( a sin u) + (a cos u) du = t a du = au ] u=t = at u= Solving for t in terms of s yields t = s/a Substituting this in the given parametric equations and using the fact that s = at ranges from to a as t ranges from to, we obtain x=acos (s/a), y=a sin (s/a) (<s<a) t Example Find Arc length of the curve r (t) = t 3 i + tj + ½ 6 t k, 1 < t < 3 Here x = t 3, y = t, z = ½ 6 t dx dt =3t, dy dz dt = 1, dt = 6 t 3 Arc length= 1 dx dt + dy dt + dz dt dt = t t dt = 1 = t 3 + t 3 1 = (3)3 + 3 (1) 3 1= = 8 (3t + 1) dt 153

154 9-Change of parameter EXAMPLE Calculate dr by chain Rule. du r = e t i + 4e -t j dr dt = et i 4e -t j dt du = u dr du = dr dt By expressing r in terms of u R = e u i + 4e -u j dr du = u eu i 8ue -u j. dt du = (et i 4e t j).(u) = u e u i 8ue -u j 154

155 3-Exact differential Lecture No -3 Exact Differential If z = f (x, y), then dz = z z dx + x y dy The result can be extended to functions of more than two independent variables. If z = f(x, y, w), dz= z x dx+ z z dy+ y w dw Make a note of these results in differential form as shown. Exercise Determine the differential dz for each of the following functions. 1. z = x + y. z = x 3 sin y 3. z = (x 1) e 3y 4. z = x + y + 3w 5. z = x 3 y w. Finish all five and then check the result. 1. dz = (x dx + y dy). dz = x (3 sin y dx + x cos y dy) 3. dz = e 3y {dx + (6x 3) dy} 4. dz = (xdx + ydy + 3wdw) 5. dz = x y (3ywdx + xwdy + xydw) Exact Differential We have just established that if z=f(x, y) dz = z z dx + x y dy We now work in reverse. Any expression dz = Pdx + Qdy, where P and Q are functions of x and y, is an exact differential if it can be integrated to determine z. P = z x and Q = z y Now P y = z Q and y x x = z x y and we know that z y x = z x y Therefore, for dz to be an exact differential P y = Q and this is the test we apply. x Example dz = (3x + 4y ) dx + 8xy dy. If we compare the right-hand side with Pdx + Qdy, then P = 3x +4y P y = 8y Q = 8xy Q x = 8y P y = Q dz is an exact differential x Similarly, we can test this one. 155

156 3-Exact differential Example dz = (1 + 8xy) dx + 5x dy. From this we find dz is not an exact differential for dz = (1 + 8xy) dx + 5x dy P = 1 + 8xy P y = 8x Q = 5x Q x = 1x P y Q dz is not an exact differential x Exercise Determine whether each of the following is an exact differential. 1. dz = 4x 3 y 3 dx + 3x 4 y dy. dz = (4x 3 y+xy 3 ) dx+(x 4 +3x y ) dy 3. dz=(15y e 3x +xy )dx+(1ye 3x +x y)dy 4. dz=(3x e y y e 3x )dx+(x 3 e y ye 3x )dy 5. dz=(4y 3 cos4x+3x cosy)dx+(3y sin4x x 3 sin y) dy. 1. Yes. Yes 3. No 4. No 5. Yes We have just tested whether certain expressions are, in fact, exact differentials and we said previously that, by definition, an exact differential can be integrated. But how exactly do we go about it? The following examples will show. Integration Of Exact Differentials dz = Pdx+Qdy where P= z and Q= z x y z = Pdx and also z = Qdy Example dz = (xy + 6x) dx + (x + y3) dy. P = z = xy + 6x z= (xy+6x)dx x z = x y + 3x + f (y) where f(y) is an arbitrary function of y only, and is akin to the constant of integration in a normal integral. Also Q = z y = x + y 3 z = (x +y 3 ) dy z = xy + y4 + F(x) where F(x) is an arbitrary function of x only. z = x y + 3x + f(y) (i) and z = x y + y4 +F(x) (ii) For these two expressions to represent the same function, then f(y) in (i) must be y4 already in (i) and F(x) in (ii) must be 3x already in (i) z = x y + 3x + y4 156

157 3-Exact differential Example Integrate dz = (8e 4x + xy ) dx + (4 cos 4y + x y) dy. Argue through the working in just the same way, from which we obtain z = e 4x + x y + sin 4y Here it is. dz = (8e 4x + xy ) dx+(4 cos4y+x y) dy P = x x = 8e4x + xy z = (8e 4x + xy ) dx z = e 4x + x y + f(y) Q = z y = 4 cos 4y + x y (i) z = (4cos 4y + x y) dy z = sin 4y + x y + F(x) (ii) For (i) and (ii) to agree, f (y) = sin 4y and F(x) = e 4x z = e 4x + x y + sin 4y Area enclosed by the closed curve One of the earliest application of integration is finding the area of a plane figure bounded by the x-axis, the curve y = f (x) and ordinates at x=x 1 and x=x. A 1 = x x 1 ydx= x f(x)dx x 1 If points A and B are joined by another curve, y = F(x) A = x f(x)dx x 1 Combining the two figures, we have A=A 1 A A= x x 1 f(x)dx x f(x)dx x 1 The final result above can be written in the form A = ydx 157

158 3-Exact differential Where the symbol O indicates that the integral is to be evaluated round the closed bound ary in the positive Example Determine the area enclosed by the graph of y = x 3 and y = 4x for x >. First we need to know the points of intersection. These are x = and x = We integrate in a an anticlockwise manner c 1 : y = x 3, limits x = to x = c : y = 4x, limits x = to x =. A = O y dx = A = 4 square units For A = O ydx = dx+ 1 x3 4xdx = x [x ] = 4 Example Find the area of the triangle with vertices (, ), (5, 3) and (, 6). The equation of OA is y = 3 5 x,ba is y = 8 x, OB is y = 3x Then A = O y dx Write down the component integrals with appropriate limits A= O ydx= 5 3 xdx+ (8 x)dx+ 5 3xdx 5 The limits chosen must progress the integration round the boundary of the figure in an anticlockwise manner. Finishing off the integration, we have A = 1 square units 158

159 3-Exact differential The actual integration is easy enough. The work we have just done leads us on to consider line integrals, so let us make a fresh start in the next frame. Line Integrals If a field exists in the xy-plane, producing a force F on a particle at K, then F can be resolved into two components.f 1 along the tangent to the curve AB at K. Fa along the normal to the curve AB at K. The work done in moving the particle through a small distance δs from K to L along the curve is then approximately F 1 δs. So the total work done in moving a particle along the curve from A to B is given by Lim F t δs = F t ds from A to B δ This is normally written AB F t ds where A and B are the end points of the curve, or as F t ds where the curve c connecting A and B is defined. C Such an integral thus formed, is called a line integral since integration is carried out along the path of the particular curve c joining A and B. I = AB F t dx = F t ds C where c is the curve y = f(x) between A(x 1, y ) and B (x, y ). There is in fact an alternative form of the integral which is often useful, so let us also consider that. Alternative form of a line integral It is often more convenient to integrate with respect to x or y than to take arc length as the variable. If F t has a component P in the x-direction Q in the y-direction then the work done from K to L can be stated as Pδx + Qδy 159

160 3-Exact differential AB F t ds = AB (P dx + Qdy) where P and Q are functions of x and y. In general then, the line integral can be expressed as I = C F t ds = (P dx + Qdy) C where c is the prescribed curve and F, or P and Q, are functions of x and y. Make a note of these results then we will apply them to one or two examples. 16

161 31-Line integral Lecture No -31 Line Integral The work done in moving the particle through a small distance δs from K to L along the curve is then approximately F 1 δs. So the total work done in moving a particle along the curve from A to B is given by Lim δ F t δs = F t ds from A to B This is normally written AB F t ds where A and B are the end points of the curve, or as F t C ds where the curve c connecting A and B is defined.such an integral thus formed, is called a line integral since integration is carried out along the path of the particular curve c joining A and B. I = AB F t dx = F t ds C where c is the curve y = f(x) between A(x 1, y ) and B (x, y ). There is in fact an alternative form of the integral which is often useful, so let us also consider that. Alternative form of a line integral It is often more convenient to integrate with respect to x or y than to take arc length as the variable. If F t has a component,p in the x-direction,q in the y-direction then the work done from K to L can be stated as Pδx + Qδy Example 1 Evaluate (x + 3y) dx from A (, 1) to B (, 5) C along the curve y = 1 + x. The line integral is of the form (P dx + Qdy) where, in this case, Q = and c C is the curve y = 1 + x. It can be converted at once into an ordinary integral by substituting for y and applying the appropriate limits of x. I = (Pdx+Qdy) = (x+3y)dx= (x+3+3x )dx C C C x = +3x+x3 =16 Now for another, so turn on. 161

162 31-Line integral Example Evaluate I = (x + y) dx + (x y )dy from A (, ) to B (, 5) along the curve y = + x. C I = (Pdx + Qdy) C P = x + y = x + + x = x + x + Q = x y = x (4+4x+x ) = (x +3x+4) Also y = + x dy = dx and the limits are x= to x=3 I = 15 for I = {(x +x+) dx (x +3x+4) dx} or (x+) dx = [ x x] 4 4 = 8 = Example 3 Evaluate I = {(x +y)dx + xydy} from O(, ) to B(1, 4) along the curve y=4x. C In this case, c is the curve y = 4x. dy = 8x dx Substitute for y in the integral and apply the limits. Then I = 9.4 for I = {(x +y) dx+xydy} C y = 4x dy = 8xdx also x + y = x + 8x = 9x ; xy = 4x 3 I= 1 {9x dx+3x 4 dx}= 1 (9x +3x 4 ) dx = 9.4 They are all done in very much the same way. Example 4 Evaluate I = C {(x + y) dx + xydy} from O(, ) to A (1, ) along the line y = and then from A (1, ) to B (1, 4) along the line x = 1. (i) OA : c1 is the line y = dy =. Substituting y = and dy = in the given integral gives. I OA = 1 x dx = x = 1 3 (ii) AB: Here c is the line x = 1 dx= I AB = 8 For I AB = 4 {(1 + y) () + ydy} = 4 y ydy = Then I = I OA +I AB = = 81 3 I=8 1 3 If we now look back to Example 3 and 4 just completed, we find that we have evaluated the same integral between the same two end points, but along different paths of integration.if we combine the two diagrams, we have where c is the curve y = 4x and c1 + c are the lines y = and x = 1. The result obtained were 4 = 8 16

163 31-Line integral I c = 9 3 and I c1+c = 81 3 Notice therefore that integration along two distinct paths joining the same two end points does not necessarily give the same results. Properties of line integrals 1. Fds = {Pdx + Qdy} C C. Fds = AB BA Fds and AB {Pdx+Qdy} = BA {Pdx+Qdy} i.e. the sign of a line integral is reversed when the direction of the integration along the path is reversed. 3. (a) For a path of integration parallel to the y-axis, i.e. x = k, dx = Pdx = I C = Qdy. C C (b) For a path of integration parallel to the x-axis, i.e. y = k, dy =. C Qdy= I C = Pdx. C 4. If the path of integration c joining A to B is divided into two parts AK and KB, then I c = I AB = I AK + I KB. 5.If the path of integration c is not single valued for part of its extent, the path is divided into two sections. y = f 1 (x) from A to K,y = f (x) from K to B. 6. In all cases, the actual path of integration involved must be continuous and singlevalued. Example Evaluate I = (x + y) dx from A(, 1) to B (, 1) along the semi-circle x +y =1 C for x >.The first thing we notice is that the path of integration c is not single-valued For any value of x, y = ± we divided c into two parts (i) y = 1 x from A to K (ii) y = 1 x from K to B As usual, I = C 1 x. Therefore, (Pdx + Qdy) and in this particular case, Q = I = Pdx = 1 (x+ 1 x ) dx + (x 1 x ) dx C 1 = 1 (x + 1 x x+ 1 x dx = 1 Now substitute x = sin θ and finish it off. I = 1 x dx 163

164 31-Line integral for I = 1 dx = cos θ dθ 1 x dx x = sin θ 1 x = cosθ Limits : x =, θ = ; x = 1, θ = I = / cosθdθ = / (1+cosθ)dθ= / sin θ θ + = Now let us extend this line of development a stage further. Example Evaluate the line integral I = O (x dx xy dy) where c comparises the three sides of the triangle joining O(, ), A (1, ) and B (, 1). First draw the diagram and mark in c 1, c and c 3, the proposed directions of integration. Do just that.the three sections of the path of integration must be arranged in an anticlockwise manner round the figure. Now we deal with each pat separately. (a) OA : c 1 is the line y = Therefore, dy =. Then I = O (x dx xy dy) for this part becomes I 1 = 1 x dx = x = 1 3 therefore I 1 = 1 3 (b) AB : for c is the line y = 1 x dy = dx. I = 1 {x dx+x(1 x)dx}= 1 (x +x x )dx = 1 (x x ) dx= x x3 3 1 = 3 I = 3 Note that anticlockwise progression is obtained by arranging the limits in the appropriate order. Now we have to determine I 3 for BO. (c) BO: c 3 is the line x = dx = I 3 = dy = I 3 = Finally, I = I 1 +I +I 3 = = 1 3 Example Evaluate O c y dx when c is the circle x +y = 4. x + y = 4 y = ± 4 x y is thus not single-valued. Therefore use y = 4 x for ALB between x = and x = and y = 4 x for BMA between x = and x =. I =

165 31-Line integral I = - 4 x dx+ - { 4 x } dx = - 4 x dx= 4 x dx = 4-4 x dx. To evaluate this integral, substitute x = sin θ and finish it off. I = 4 165

166 3-Examples Lecture No -3 Examples Example Evaluate I = O {xydx+(1+y )dy} where c is the boundary of the rectangle joining A(1,), B (3, ), C(3, ), D (1, ). First draw the diagram and insert c 1,c,c 3,c 4. That give Now evaluate I 1 for AB; I for BC; I 3 for CD; I 4 for DA; and finally I. I 1 = ; I =4 3 ; I 3 = 8 I 4 = 4 3 ; I = 8 Here is the complete working. I = O {xydx + (1 + y ) dy} (a) AB: c 1 is y = dy = I 1 = (b) BC: c is x = 3 dx = I = (1+y )dy = y + y3 3 = 4 3 I = 4 3 (c) CD : c 3 is y = dy = I 3 = 1 xdx = [x ] 1 = 8 I 3 = (d) DA : c 4 is x = 1 dx = I 4 = (1+y ) dy= y + y3 3 = 4 3 Finally I = I 1 + I + I 3 + I 4 = = 8 I = 8 Remember that, unless we are directed otherwise, we always proceed round the closed boundary in an anticlockwise manner. Line integral with respect to arc length We have already established that I = AB F t ds = {Pdx+Qdy} AB where F t denoted the tangential force along the curve c at the sample point K(x,y). The same kind of integral can, of course, relate to any function f(x,y) which is a function of the position of a point on the stated curve, so that I = f(x, y) ds. C This can readily be converted into an integral in terms of x: f(x,y)dx = f(x,y) ds I = C where ds C dx dx dx = 1 + dy dx f(x,y) dx= x f(x,y) 1+ dy C x 1 dx dx (1) 166

167 3-Examples Example Evalaute I = (4x+3xy)ds where c is the straight line joining O(,) to A (1,). C c is the line y = x dy dx = ds dx = 1 + dy dx = 5 I = x=1 (4x+3xy)ds = 1 (4x+3xy)( 5) dx. But y = x x= for I = 1 (4x+6x )( 5) dx = 5 1 (x+3x ) dx = 4 5 Parametric Equations When x and y are expressed in parametric form, e.g. x = y (t), y = g(t), then ds dt = dx dt + dy dt ds = dx dt + dy dt dt I= f(x,y)ds= t f(x,y) C t 1 dx dt + dy dt () Example Evaluate I = O 4xyds where c is defined as the curve x = sin t, y = cos t between t= and t= 4. We have x = sin t ds dt = I for ds dt = dx dt + dy dt dx dt = cos t, y = cos t dy dt = cos t+sin t =1 = sin t I = t t 1 f(x,y) = /4 dx dt + dy dt dt 4 sin t cos t dt = /4 = cos t /4 = 1 sin t dt Dependence of the line integral on the path of integration We know that integration along two separate paths joining the same two end points does not necessarily give identical results.with this in mind, let us investigate the following problem. 167

168 3-Examples EXAMPLE Evaluate I = O C {3x y dx + x 3 y dy} between O (, ) and A (, 4) (a) along c 1 i.e. y = x (b) along c i.e. y = x (c) along c 3 i.e. x = from (,) to (,4) and y = 4 from (,4) to (,4). (a).first we draw the figure and insert relevant information. I = {3x y dx + x 3 ydy} C The path c 1 is y = x dy = x dx I 1 = {3x x 4 dx+x 3 x xdx}= (3x 6 + 4x 6 ) dx = [x 7 ] = 18 I 1 = 18 (b) In (b), the path of integration changes to c, i.e. y = x So, in this case,for with c, y = x dy = dx I = (3x 4x dx + x 3 x dx} = x 4 dx= 4[x 5 ] = 18 I = 18 (c) In the third case, the path c 3 is split x = from (,) to (, 4), y = 4 from (, 4) to (, 4) Sketch the diagram and determine I 3. from (,) to (,4) x= dx= I 3a = from (,4) to (,4) y=4 dy= I 3b =48 48x dx = 18 I 3 = 18 In the example we have just worked through, we took three different paths and in each case, the line integral produced the same result. It appears, therefore, that in this case, the value of the integral is independent of the path of integration taken. We have been dealing with I = {3x y dx+x 3 ydy} C 168

169 3-Examples On reflection, we see that the integrand 3x y dx + x 3 y dy is of the form Pdx+Qdy which we have met before and that it is, in fact, an exact differential of the function z = x 3 y z, for x = 3x y and z y = x3 y This always happens. If the integrand of the given integral is seen to be an exact differential, then the value o the line integral is independent of the path taken and depends only on the coordinates of the two end points. 169

170 33-Examples Lecture No -33 Examples Example Evaluate I = {3ydx + (3x+y)dy} from A(1, ) to B (3, 5). C No path is given, so the integrand is doubtless an exact differential of some function z = f (x,y). In fact P y = 3 = Q. We have already dealt with the integration of exact x differentials, so there is no difficulty. Compare with I = {P dx + Q dy}. C P = z x = 3y z = 3ydz=3xy+f(y) (i) Q = z = 3x + y z = (3x+y) dy = 3xy + y +F(x) (ii) y For (i) and (ii) to agree f(y) = y ; F(x) = Hence z = 3xy + y {3ydx + (3x+y)dy}= (3,5) I = C Example (1,) d(3xy+y )=[3xy+y ] (3,5) Evaluate I = {(x +ye x )dx+(e x +y)dy} between A (, 1) and B (1, ). C As before, compare with {Pdx+Q dy}. C P = z x =x +ye x z = x3 3 + yex +f (y) Q = z y =ex +y z = ye x + y + F(x) For these expressions to agree, f(y) = y x3 ; F(x) = 3 Then I = x yex + y (1,) = e (,1) So the main points are that, if (Pdx+Qdy) is an exact differential (a) I = (Pdx + Qdy) is independent of the path of integration C (b) I = O (P dx + Q dy) is zero. C If I = {P dx + Q dy} and (Pdx + Qdy) is an exact differential, C Then I c1 = I c I c1 + I c = Hence, the integration taken round a closed curve is zero, provided (Pdx+Q dy) is an exact differential. (1,) If (P dx + Q dy) is an exact differential, O (P dx + Q dy) = = (45+5) (6+4) = 6 17

171 33-Examples Exact differentials in three independent variables A line integral in space naturally involves three independent variables, but the method is very much like that for two independent variables. dz = Pdx + Q dy + R dw is an exact differential of z = f(x, y, w) if P y = Q x ; P w = R x ; R y = Q w If the test is successful, then (a) (P dx + Q dy + R dw) is independent of the path of integration. C (b) O (P dx + Q dy + R dw) is zero. C Example Verify that dz=(3x yw+6x)dx+(x 3 w 8y)dy+(x 3 y+1) dw is inexact differential and hence evaluate dz from A (1,, 4) to B (, 1 3). C First check that dz is an exact differential by finding the partial derivatives above, when P = 3x yw + 6x; Q = x 3 w 8y; and R = x 3 y + 1 P y = 3x w ; Q x = 3x w P y = Q x P w = 3x y ; R x = 3x y P w = R x R y = x3 ; Q w = x3 R y = Q w dz is an exact differential Now to find z. P = z x ; Q = z y ; R = z w z x =3x yw+6x z= (3x yw+6x)dx = x 3 yw+3x +f(y)+f(w) z y =x3 w 8x z = (x 3 w 8y)dy = x 3 yw 4y +g(x)+f(w) z w =x3 y+1 z = (x 3 y+1)dw = y 3 yw+w+f(y)+g(x) For these three expressions for z to agree f(y) = 4y ; F(w) = w; g(x) = 3x z = x 3 yw + 3x 4y + w I = [x 3 yw + 3x 4y +w] (,1,3) (1,,4) for I = [x 3 yw + 3x 4y +w] (,1,3) (1,,4) = ( ) ( )=36 The extension to line integrals in space is thus quite straightforward. Finally, we have a theorem that can be very helpful on occasions and which links up with the work we have been doing. It is important, so let us start a new section. 171

172 33-Examples Green s Thorem Let P and Q be two function of x and y that are finite and continuous inside and the boundary c of a region R in the xy-plane.if the first partial derivatives are continuous within the region and on the boundary, then Green s theorem states that. P R y Q x dx dy = O (P dx+ Q dy) C That is, a double integral over the plane region R can be transformed into a line integral over the boundary c of the region and the action is reversible. Let us see how it works. EXAMPLE Evaluate I = O {(x y)dx + (y+x)dy} around the boundary c. the ellipse C x + 9y = 16. The integral is of the form I = O {P dx + Q dy) where P = x y P Q = 1 and Q = y + x C y x = 1. I = P R y Q x dxdy= dy= dx dy R ( 1 1)dx R But dx dy over any closed region give the area of the figure. R In this case, then, I = 4 where A is the area of the ellipse x +9y = 16 i.e. x 16 +9y 16 = 1 a = 4; b = 4 3 A = 16 3 I = A = 3 3 To demonstrate the advantage of Green s theorem, let us work through the next example (a) by the previous method, and (b) by applying Green s theorem. Example Evaluate I = O {(x+y) dx+(3x y) dy} taken in anticlockwise manner round the triangle C with vertices at O (,) A (1, ) B (1, ). I = O {(x + y) dx + (3x y) dy} C 17

173 33-Examples (a) By the previous method There are clearly three stages with c 1,c,c 3. Work through the complete evaluation to determine the value of I. It will be good revision. When you have finished, check the result with the solution in the next frame. I = (a) (i) c 1 is y = dy = I 1 = 1 x dx = [x ] 1 = 1 I 1 = 1 (ii) c is x = 1 dx = (3 y) dy=[3y y ] = I = 1 (iii) c 3 is y = x dy = dx I 3 = {4x dx + (3x 4x) dx} 1 = x dx = [x ] = 1 I 3 = I = I 1 +I +I 3 = 1++( 1) = I = I = Now we will do the same problem by applying Green s theorem, so more (b) By Green s theorem I = O C {(x + y) dx + (3x y) dy} P = x + y Q = 3x y I = P R x Finish it off. I = y Q P y =1; Q x = 3 dx dy For I = (1 3) dx dy= dx dy = A R R = the area of the triangle = 1 = I = Application of Green s theorem is not always the quickest method. It is useful, however, to have both methods available. If you have not already done so, make a note of Green s theorem. R P y Q x dx dy = O C (P dx + Q dy) Note: Green s theorem can, in fact, be applied to a region that is not simply connected by arranging a link between outer and inner boundaries, provided the path of integration is such that the region is kept on the left-hand side. 173

174 34-Examples Example Lecture -34 Examples Evaluate the line integral I = O {xy dx + (x y) dy} round the region bounded by C the curves y = x and x = y by the use of Green s theorem. Points of intersection are O(, ) and A(1, 1). O C {xy dx + (x y) dy} I = O C {Pdx+ Qdy}= R P = xy P y I = R = 1 I = 1 P y Q x dx dy = x; Q = x y Q x = (x ) dx dy = 1 (x ) [ y ] x x dx y= x y=x (x ) ( ) (x ) dy dx x x dx = 1 = 5 x5/ 1 4 x4 4 3 x3/ + 3 x3 1 = 31 6 In this special case when P=y and Q= x so P y Green s theorem then states R {1 ( 1)} dx dy = O C R dx dy = O C (x 3/ x 3 x 1/ + x ) dx = 1 and Q x = 1 (P dx+q dy) i.e. (y dx x dy) = O (x dy y dx) C Therefore, the area of the closed region A = dy = R dx 1 O C Example Determine the area of the figure enclosed by y = 3x and y = 6x. Points of intersection : 3x = 6x x = or Area A = 1 O C (x dy y dx) (x dy y dx) We evaluate the integral in two parts, i.e. OA along c 1 and AO along c (xdy ydx) (xdy ydx) A= c1 + (along OA) c (along OA) I 1 : c 1 is y = 3x dy = 6x dx = I 1 +I 174

175 34-Examples I 1 = (6x dx 3x dx) = Similarly, for c is y = 6x dy = 6 dx I = (6x dx 6x dx) = I = I = I 1 + I = 8 + = 8 A = 4 square units 3x dx = [ x 3 ] = 8 I 1 = 8 Example Determine the area bounded by the curves y = x 3, y = x 3 x = for x >. Here it is y = x 3 ; y = x 3 + 1; x = Point of intersection x 3 = x x 3 = 1 x = 1 Area A = 1 O (x dy y dx) A = O C C (a) OA : c 1 is y = x 3 dy = 6x dx I 1 = c1 (xdy ydx)= 1 (6x3 dx x 3 dx) = 1 I 1 = 1 (b) AB: c is y = x dy = 3x dx I = 1 {3x3 dx (x 3 + 1) dx} = (x dy y dx) 4x3 dx = [ x4 ] 1 1 (x3 1) dx = x 4 x 1 = 1 = 1 1 = and the axis I = 1 (c) BO: c 3 is x = dx = I 3 = y= y=1 (xdy ydx)= I 3 = A = I = I 1 + I + I 3 = = 11 A = 3 4 square units Revision Summary Properties of line integrals Sign of line integral is reversed when the direction of integration along the path is reversed. Path of integration parallel to y-axis, dx = I c = c Q dy. Path of integration parallel to x-axis, dy = I c = c P dx. Path of integration must be continuous and single-valued. Dependence of line integral on path of integration. In general, the value of the line integral depends on the particular path of integration. Exact differential If P dx + Q dy is an exact differential 175

176 34-Examples (a) P y = Q x (b) I = c (c) I = O C (P dx + Q dy) is independent of the path of integration (P dx + Q dy) is zero. Exact differential in three variables. If P dx + Q dy + R dw is an exact differential (a) P y = Q x ; P w = R x ; R y = Q w (b) c (P dx + Q dy + R dw) is independent of the path of integration. (c) O C (P dx + Q dy + R dw) is zero. Green s theorem O (P dx+q dy)= C R P y Q dx dy and, for a simple closed curve, x O (x dy y dx) = dx dy = A C R where A is the area of the enclosed figure. Gradient of a scalar function Del operator is given by = i x + j y + k z φ = grad φ = i x + j y + k z φ = i φ x + j φ y + k φ z gradφ = φ = φ x i + φ y j + φ z k Div (Divergence of a vector function) If A = a 1 i + a j + a 3 k then div A =.A = i x + j y + k z.(a 1i + a j + a 3 k) div A =.A = a 1 x + a y + a 3 z Note that (a) the grad operator acts on a scalar and gives a vector (b) the div operator. acts on a vector and gives a scalar. Example If A = x yi xyzj + yz k then 4 Div A =.A = ( x y) ( xyz) + ( yz ) = xy xz + yz x y z Example 176

177 34-Examples If A = x yi (xy +y 3 )j+3y z k determine.a i.e. div A. A= x yi (xy + y 3 z)j + 3y z k.a= a x x + a y y + a z z = 4xy (xy+3y z)+6y z= 4xy 4xy 6y z+6y z = Such a vector A for which.a = at all points, i.e. for all values of x, y, z, is called a solenoid vector. It is rather a special case. Curl (curl of a vector function) The curl operator denoted by, acts on a vector and gives another vector as a result. If A= a 1 i + a j + a 3 k then curl A= A. i.e. curl A= A= i x + j y + k z (a 1i + a j + a 3 k) i j k = x y z a 1 a a 3 A= i a 3 y a z +j a 1 z a 3 x + k a x a 1 y Curl A is thus a vector function. Example If A=(y 4 x z )i+(x +y )j x yzk,determine curl A at the point (1,3, ). i j k Curl A = A = x y z y4 x z x +y x yz Now we expand the determinant A= i y ( x yz) z (x +y ) j x ( x yz) z (y4 x z ) + k x (x + y ) y (y4 x z ) A=i{ x z} j{ xyz+x z}+k(x 4y 3 }. At (1, 3, ), A = i () j (1 4) + k ( 18) = i 8j 16k Example Determine curl F at the point (,,3) given that F=ze xy i+xycosyj+(x+y)k. i j k In determine form, curl F = F = x y z ze xy xzcosy x+y Now expand the determinant and substitute the values for x, y and z, finally obtaining curl F = i{ x cos y} j{1 e xy }+ k ({z cos y xze xy } At(,,3) F= i( 4) j(1 1)+k(6 1) = i 6k = (i + 3k) Summary of grad, div and curl 177

178 34-Examples (a) Grad operator acts on a scalar field to give a vector field. (b) Div operator. Acts o a vector field to give a scalar field. (c) Curl operator acts on a vector field to give a vector field. (d) With a scalar function φ (x,y,z) Grad φ = φ = φ x i+ φ y j+ φ z k (e) With a vector function A=a x i+a y j+a z k (i) div A =.A= a x x + a y y + a z z (ii) Curl A = A = i j k x y z a x a y a z Multiple Operations We can combine the operators grad, div and curl in multiple operations, as in the examples that follow. EXAMPLE If A = x yi + yz 3 j zx 3 k Then div A=.A = i x + j y + k z.(x yi + yz 3 j zx 3 k) = xy+z 3 + x 3 = φ (say) Then grad (div A) = (.A) = φ x i + φ y j + φ z k = (y+3x )i+(x)j+(3z )k i.e., grad div A = (.A)= (y+3x )i+xj+3z k Example If φ = xyz y z + x z determine div grad φ at the point (, 4, 1). First find grad φ and then the div of the result. div grad φ =.( φ) We have φ = xyz y z + x z grad φ = φ = φ x i + φ y j + φ z k =(yz+xz )i+(xz 4yz)j+(xy y +x z) k div grad φ=.( φ) = z 4z+x At (,4,1), div grad φ =.( φ) = = 6 grad φ = φ x i + φ y j + φ z k Then div grad φ =.( φ)= i x + j y + k z. φ x i+ φ y j+ φ z k = φ x + φ z + φ z div grad φ=.( φ) = φ x + φ y + φ z 178

179 34-Examples Example If F = x yzi + xyz j + y zk determine curl F at the point (, 1, 1).Determine an expression for curl F in the usual way, which will be a vector, and then the curl of the result. Finally substitute values. Curl curl F= ( F) = i+j+6k i j k = (yz xyz)i+x x y z yj+(yz x z)k x yz xyz y z i j k Then Curl Curl F = x y z = z i ( xz y+xy)j+(xy z+xz)k yz xyz x y yz x z At (, 1, 1), curl cul F= ( F) = i + j + 6k For curl F = Two interesting general results (a) Curl grad φ where φ is a scalar grad φ = φ x i + φ y j + φ z k i j k curl grad φ = x y z φ φ φ x y z = i φ y z φ z x j φ z x φ x z + k φ x y φ y x = curl grad φ = ( φ) = (b) Div curl A where A is a vector. A = a x i + a y j + a z k i j k a z curl A = A= = i x y z y a y z j a z x a x z a x a y a z Then div curl A =.( A) = i x + j y + k z. ( A) = a z x y a y z x a z x y + a x y z a y z x a x y z = div curl A =. ( A) = (c) Div grad φ where φ is a scalar. grad φ = φ x i + φ y j + φ z k + k a y x a x y Then div grad φ =.( φ)= i x + j y + k z. φ x i+ φ y j+ φ z k = φ x + φ z + φ z div grad φ=.( φ) = φ x + φ y + φ z This result is sometimes denoted by φ. So these general results are 179

180 34-Examples (a) curl grad φ = ( φ) = (b) div curl A =.( A) = (c) div grad φ=.( φ)= φ x + φ y + φ z 18

181 35-Definite intagrals Lecture No -35 Definite Integrals Definite integral for sin n x and cos n x, x / 1 1 sinx 1 sin sin xdx = (1 cos ) x dx = x = = 4 sin 1 xdx = 1 1 sinx 1 sin cos xdx = (1 cos ) + x dx = x + = + = 4 cos 1 xdx = sin xdx = sin xsin xdx = (1 cos x)sin xdx = sin xdx + cos x( sin x) dx 3 cos x = cos x + = cos + cos + cos cos = = cos xdx = cos x cos xdx = (1 sin x) cos xdx = cos xdx sin x(cos x) dx 3 sin x = sin x = sin sin sin sin = = cosx 1 sin xdx = (sin x) dx = dx = (1 cos x + cos x) dx 4 sin cos4 1 3 os4 x c x = (1 cos ) ( cos ) 4 x + dx = x dx sin sin x = x sin x+ = sin xdx = 4 so sin 4 31 xdx = 4 181

182 35-Definite intagrals 4 1+ cosx 1 cos xdx = (cos x) dx = dx = (1 + cos x + cos x) dx 4 cos sin xdx = and cos xdx = sin xdx = and cos xdx = sin xdx = and cos xdx = sin xdx = and cos xdx = sin xdx = and cos xdx = sin xdx = and cos xdx = Wallis Sine Formula When n is even 1 1+ cos4 1 3 os4 x c x = (1 cos ) ( cos ) 4 + x + dx = x dx sin sin x = x+ sin x+ = sin xdx = 4 So cos 4 31 xdx = 4 n n 1 n 3 n 5 n sin xdx = n n n 4 n When n is odd n n 1 n 3 n 5 n cos xdx = n n n 4 n

183 35-Definite intagrals sin xdx = and cos xdx = sin xdx = and cos xdx = Integration By Parts Example Evaluate UVdx U V dx Vdx. du = dx dx x lnx dx x lnx dx= lnx xdx [ x dx. d dx (ln x)] dx (We are integrating by parts) x = ln x( )- ( x 1 )( x )dx=( x ) ln x- ( x x 1 )d x = ( ) ln x- ( x ) Example Evaluate x sinx dx x sinx dx =x sin xdx [ sinx dx. d dx = x(-cosx)- (-cosx)(1)dx= -x(cosx ) + Line Integrals Let a point p on the curve c joining A and B be denoted by the position vector r with respect to origin O. If q is a neighboring point on the curve with position vector r+dr, then PQ = r The curve c can be divided up into many n such small arcs, approximating to dr 1, dr, dr 3,. dr p, so that AB n p= 1 dr p (x)] dx (We are integrating by parts) cosx d x = -x(cosx)+sin x where dr p is a vector representing the element of the arc in both magnitude and direction. If dr, then the length of the curve AB= dr. Scalar Field If a scalar field V(r) exists for all points on the curve, n the V() r drp with dr, defines the line integral p= 1 of V i.e line integral = V() r dr. c We can illustrate this integral by erecting a continuous Ordinate to V(r) at each point of the curve V( r) dr is then represented by the area of the curved surface between the ends A and B the curve c. To evaluate a line integral, the integrand is expressed in terms of x, y,z with dr =dx + dy j + dz k In practice, x, y and are often expressed in terms of parametric equation of a fourth variable (say u), i.e. x = x(u) ; y = y(u) ; z = z(u). From these, dx, dy and dz can be written in terms of u and the integral evaluate in terms of this parameter u. c c 183

184 36-Scalar field Lecture No -36 Scalar Field Scalar Field If a scalar field V(r) exists for all points on the curve, n the V() r drp with dr, defines the line integral p= 1 of V i.e line integral = V() r dr. c We can illustrate this integral by erecting a continuous Ordinate to V(r) at each point of the curve V( r) dr is then represented by the area of the curved surface between the ends A and B the curve c. To evaluate a line integral, the integrand is expressed in terms of x, y,z with dr =dx + dy j + dz k In practice, x, y and are often expressed in terms of parametric equation of a fourth variable (say u), i.e. x = x(u) ; y = y(u) ; z = z(u). From these, dx, dy and dz can be written in terms of u and the integral evaluate in terms of this parameter u. Example If V=xy z, evaluate V( r) dr along the curve c having parametric equations c x = 3u; y=u ;z=u 3 between A(,,) and B(3,,1) V= xy z = (3u)(4u 4 )(u 3 )=1u 8 dr= dxi+ dy j+ dz k dr= 3du i +4udu j +3u du k for x = 3u ; dx = 3du ; y = u dy = 4u du ; z = u 3 dz =3u dz Limiting : A(,,) corresponds to B(3,,1) corresponds to u A(,,) u= ; B(3,,1) u = u u u 4 36 V ( r) dr = 1u (3 i +4u j +3u k)du= 36 i + 48 j + 36 = 4i + j + k c Example If V = xy + y z Evaluate V( r) dr along the curve c defined by x= t ; y = t ; z= t+5 c between A(,,5) and B(4,4,7). As before, expressing V and dr in term of the parameter t. since V=xy+y z = (t )(t)+(4t )(t+5) = 6t 3 + t. x = t dx = t dt y = t dy = dt } z = t+5 dz = dt dr = dxi + dy j + dzk C = t dt i + d t j + dtk Vdr = (6t3 +t )( t i + j+k) dt C c 184

185 36-Scalar field Limits: A (,, 5) t = ; B (4, 4, 7) t = C Vdr = (6t 3 +t )( t i + j +k) dt Vdr = {6t 4 +t 3 )i+(6t 3 +t )j C +(3t 3 +1t )k}dt. = 8 (444i + 9j + 145k) 15 Vector Field If a vector field F(r) exist for all points of the curve c, then for each element of arc we can form the scalar product F.dr. Summing these products for all elements of arc, we have n p= 1 Fdr. p T he line integral of F(r) fr along the stated curve = F.dr. C In this case, since om A to B F.d r is a scalar product, then the line integral is a scalar. To evaluate the line integral, F and d r are expressed in terms of x,y,z, and the curve in parametric form. We have F= F 1 i + F j +F 3 k And dr = dx i +dy j +dz k Then F.dr = (F 1 i + F j +F 3 k).( dx i +dy j +dz k) = (Fdx 1 + Fdy +F3dz ) Now for an example to show it in operation. Example If F(r) = x y i + xz j + yz k, Evaluate Fdr. between A(,,) and B(4,,1) along the curve c having parametric equations x=4t ; y-t ; z = t 3 Expressing everthing in terms of the parameter t, we have dx = 4 dt ; dy =4tdt ; dz = 3t dt x y = (16t )(t ) = 3 t 4 x = 4t dx = 4 dt xz = (4t)(t 3 ) = 4 t 4 y = t dy = 4t dt yz = (4 t )( t 3 ) = 4 t 5 z = t 3 dz = 3t dt F = 3 t 4 i + 4 t 4 j 4 t 5 k dr = 4dt i + 4t dt j + 3t k Then F. d r = (3t 4 i+4t 4 j 4t 5 k). (4dt i + 4t dt j + 3t dt k) = (18t 4 +16t 5 + 1t 7 ) dt Limits: A(,,) t = ; B (4,, 1) t = 1 F. d r = (18t t 5 + 1t 7 )dt = 18 5 C c t t t8 = 18 5 c =

186 36-Scalar field Example If F(r ) = x y i + yzj + 3z x k Evaluate F. d r between A(,,) and B(1,,3) C B (1,, 3) (a) along the straight line c 1 from (,, ) to (1,, ) then c from (1,, ) to (1,, ) and c 3 from (1,, ) to (1,, 3) (b) along the straight line c 4 joining (,, ) to (1,, 3). W e first obtain an expression for which is F. d r = (x y i + yzj + 3z x k). (dx i+ dy j + dz k) F. d r = x y dx + yz dy + 3z x dz F.d r = x ydx + yzdy+ 3z xdz Here the integration is made in three sections, along c 1, c and c 3. (i) c 1 : y =, z =, dy =, dz = F. d r=++ = C 1 (ii) c : The conditions along c are c : x = 1, z =, dx =, dz = F.dr = + + = C (iii) c 3 : x = 1, y =, dx =, dy = C 3 F. dr = z dz = 7 Summing the three partial results ( 1,, 3) (,, ) F.d r = = 7 F.dr F. d r = 7 c 1+ c + c 3 If t taken as the parameter, the parametric equation of c are x = t; y = t; z = 3t (,, ) t =, (1,, 3) t = 1 and the limits of t are t = and t = 1 F = t 3 i + 1t j + 7t 3 k d r = dx i +dy 1 j +kdz = dt i + dt j +3dt k 1 F. d r = = (t 3 +4t +81t 3 )dt (t3 i +1t 3 j+7t 3 k). (i+j+3k) dt = (83t t ) dt = 83 t t3 = = 8.75 C 4 186

187 36-Scalar field So the value of the line integral depends on the path taken between the two end points A and B (a) F. d r via c 1, c and c 3 = 7 (b) F. d r via c 4 = 8.75 Example Evaluate Fdv where V is the region bounded by the planes x =,y =, z = and v x + y =, and F = z i +y k. To sketch the surface x + y + z =, note that when z =, x+y= i.e. y = x when y =, x+z= i.e. z = x when x =, y+z= i.e. z = y Inserting these in the planes x =, y =, z = will help. The diagram is therefore. So x + y + z = cuts the axes at A(1,,); B (,, ); C (,, ). Also F = zi + yk; z = x y = (1 x) y v FdV= = = 1 1 ( 1- x ) ( 1- x ) - y ( 1 - x ) [z i +yzk ] z= ( 1 - x ) - y 1 ( 1 - x) (xi+y k)dzdydx z= dydx {[4(1 x) 4(1 x)y+y ]i +[(1 x)y y ]k}dydx 1 v FdV = (i + k ) 3 187

188 37-Examples Lecture No -37 Examples Example Evaluate v FdV where F=i+zj+yk and V is the region bounded by the planes z =, z = 4 and the surface x +y = 9. It will be convenient to use cylindrical polar coordinates (r, θ, z) so the relevant transformations are x = rcosθ; y = rsinθ z = z; dv=rdrdθ dz Then v FdV= v (i+zj+yk)dxdydz Changing into cylindrical polar coordinates with appropriate change of limits this becomes v FdV= (i+zj+rsinθk)rdzdrdθ= θ= r= z= θ= r=[zi+z 4 rdrdθ j+rsinθzk]z= 3 3 = (8i+16j+4rsinθk)rdrdθ= 4 (ri+4rj+r sin θk) drdθ= 4 r i+r j+ r3 3 3 sinθ k dθ = 4 (9i+18j + 9 sin θk) dθ= 36 = 36 {(i + 4j k) ( k)}= 7 (i + j) (i+j+sinθk)dθ= 36 [θi + θj cosθk] Scalar Fields A scalar field F = xyz exists over the curved surface S defined by x +y = 4 between the planes z = and z = 3 in the first octant. Evaluate S F ds over this surface. We have F = xyz S: x +y 4 =, z = to z = 3 ds = ^n ds where ^n = S S Now S = S x i+ S y j+ S z k = xi+yj S = 4x +4y = x +y = 4 = 4 ^n S xi + yj = = S ds = ^nds = xi+yj ds S FdS= S F ^nds= 1 S xyz(xi+yj)ds = 1 S (x yzi+xy zj)ds (1) We have to evaluate this integral over the prescribed surface. Changing to cylindrical coordinates with r = x = cos θ; y = sin θz = z; ds = dθdz x yz = (4cos θ)(sinθ)(z)= 8 cos θ sinθz xy z = (cosθ) (4sin θ) (z)=8cosθsin θz Then result S FdS = 1 S (x yzi+xy zj)ds becomes. 188

189 37-Examples S FdS = 8 / 3 / (cos θsinθi+cosθsinθj)zdzdθ=4 / 3 (cos θsinθi+cosθsin θj)zdz dθ / = 4 (cos θsinθi+cosθsin θj)z 3 dθ= 4 (cos θsinθi+cosθsin θj)9dθ = 36 cos3 θ 3 i + sin3 / θ 3 j = 1 (i + j) Vector Field A vector field F=yi+j+k exists over a surface S defined by x +y + z = 9 bounded by x =, y =, z = in the first octant, Evaluate S F.dS over the surface indicated. ds = ^nds ^ ; n = S S S : x + y + z 9 = S = S x i + S y j+ S z k=xi+yj+zk S = 4x +4y +4z = x +y +z = 9 = 6 ^n = 1 6 (xi + yj + zk) = 1 3 (xi + yj + zk) S F.dS= S F. ^nds= S (yi+j+k). 1 3 (xi+yj+zk)ds= 1 3 S (xy + y + z) ds Before integrating over the surface, we convert to spherical polar coordinates. x = 3 sin φ cos θ; y=3 sin φ sin θ z = 3 cos φ; ds=9sinφdφdθ Limits of φ and θ are φ = to ; θ = to S F.dS = 1 3 S (xy + y + z) ds xy = 3 sinφ cosθ. 3 sinφ sinθ = 9 sin θ cosθ y =. 3 sinφ sinθ = 6 sinφ sinθ z = 3 cosφ ds = 9 sinφ dφ dθ Putting these values we get S F.dS= 1 3 / / / / (9sin φsinθcosθ+6sinφsinθ+3cosφ)9sinφdφdθ = 9 (3sin3 φsinθcosθ+sin φsinθ+sinφcosφ)dφdθ As we know that / sin 3 φ dφ = 3 / also sin φ dφ = 1 So we get / by Wallis Formula S F.dS=9 sinθcos θ+ sinθ+1 dθ = 9 sin θ cosθ+θ / = 9 (1 + 4 ) ( + ) 189

190 37-Examples = = Example Evaluate S F.dS where F = yj + zk and S is the surface x + y = 4 in the first two octants bounded by the planes z =, z = 5 and y =. S : x + y 4 = ^n = S S S = S x i + S y j + S k = xi + yi z S = 4x + 4y = x + y = 4 = 4 ^n = S xi + yj = S 4 = 1 (xi + yj) S F.dS = S F. ^n ds S F.^n ds = 1 S (y ) ds= S y ds This is clearly a case for using cylindrical polar coordinates. x = cos θ; y = sin θ z = z; ds = dθ dz sin θ dθ dz S F.dS = S y ds = S 4sin θdθdz = 8 S Limits: θ = to θ = ; z = to z = 5 S F.dS=4 5 z= (1 cos θ) dθdz = 4 5 θ= sin θ θ dz= 4 5 dz= 4 [z] 5 = Conservative Vector Fields In general, the value of the integral C F.dr between two stated points A and B depends on the particular path of integration followed. If, however, the line integral between A and B is independent of the path of integration between the two end points, then the vector field F is said to be conservative. It follows that, for a closed path in a conservative 19

191 37-Examples field, O C F.dr =. For, if the field is conservative, C1 (AB) F.dr = c (AB) F.dr But c (BA) F.dr c (AB) F.dr Hence, for the closed path AB c1 + BA c, O F.dr = C1 (AB) F.dr + c (AB) F.dr= C1 (AB) F.dr c (AB) F.dr = C1 (AB) F.dr c1 (AB) F.dr = 191

192 38-vector field Lecture No -38 Vector Field Conservative Vector Fields In general, the value of the integral C F.dr between two stated points A and B depends on the particular path of integration followed. If, however, the line integral between A and B is independent of the path of integration between the two end points, then the vector field F is said to be conservative. It follows that, for a closed path in a conservative field, O C For, if the field is conservative, C1 (AB) F.dr = c (AB) F.dr But c (BA) F.dr c (AB) F.dr Hence, for the closed path AB c1 + BA c, O F.dr = C1 (AB) F.dr + c (AB) F.dr = C1 (AB) F.dr c (AB) F.dr= C1 (AB) F.dr c1 (AB) F.dr = Note that this result holds good only for a closed curve and when the vector field is a conservative field. Now for some examples Example F.dr =. If F = xyzi + x zj + x yk, evaluate the line integral F.dr between A(,,) and B(,4,6) (a) along the curve c whose parametric equations are x = u, y = u, z = 3u (b) along the three straight lines c 1 :(,,) to (,, ); c : (,, ) to (, 4, ); c 3 : (, 4, ) to (, 4, 6). Hence determine whether or not F is a conservative field. First draw the diagram. (a) F = xyzi + x zj + x yk x = u; y = u; z = 3u dx = du; dy = udu; dz = 3du. F.dr=(xyzi+x zj+x yk).(dxi+dyj+dzk) = xyz dx + x z dy + x y dz Using the transformation shown above, we can now express F.dr in terms of u. for xyzdx=(u)(u )(3u)du = 6u 4 du x zdy = (u)(3u)(u)du = 6u 4 du x ydz = (u )(u )3du = 3u 4 du F.dr = 15u 4 du The limits of integration in u are u = to u = C F.dr = 15u4 du = [3u 5 ] = 96 19

193 38-vector field (b) The diagram for (b) is as shown. We consider each straight line section in turn. F.dr = (xyz dx + x zdy+x ydz) c 1 : (,,) to (,,); y =, z =, dy =, dz = C1 F.dr = + + = In the same way, we evaluate the line integral along c and c 3. C1 F.dr = ; F.dr= (xyzdx+x zdy+x ydz) c : (,,) to (,4,); x =, z =, dx=, dz = C F.dr = ++ = C F.dr = c 3 : (,4,) to (,4,6); x =, y = 4, dx=, dy = C3 F.dr = dz=[16z]6 = 96 C3 F.dr = 96 Collecting the three results together c1 +c +c 3 F.dr = c1 +c +c 3 F.dr = 96 In this particular example, the value of the line integral is independent of the two paths we have used joining the same two end points and indicates that F is a conservative field. It follows that i j k curl F = x y z xyz x z x y = (x x )i (xy xy)j+(xz xz)k = curl F = So three tests can be applied to determine whether or not a vector field is conservative. They are (a) O F.dr = (b) curl F = (c) F = grad V Any one of these conditions can be applied as is convenient. Divergence Theorem (Gauss theorem) For a closed surface S, enclosing a region V in a vector field F V div F dv = S F.dS In general, this means that the volume integral (tripple integral) on the left-hand side can be expressed as a surface integral (double integral) on the right-hand side. Example 193

194 38-vector field Verify the divergence theorem for the vector field F = x i + zj + yk taken over the region bonded by the planes z =, z =, x =, x = 1, y =, y = 3. dv = dx dy dz we have to show that V div F dv = S F.dS (a) To find V div F dv div F =.F = x i + y j + z k. (x i + zj + yk) = x (x )+ y (z)+ z (y)=x++=x V div F dv= V x dv= V xdzdydx V div F dv = 1 xdzdydx = 1 3 [ xz] dy dx = 1 3 (b) to find S F.dS i.e. S F.^n ds The enclosing surface S consists of six separate plane faces denoted as S 1, S,.. S 6 as shown. We consider each face in turn. F = x i + zj + yk (i) S 1 (base): z = ; ^n = k (outwards and downwards) F = x i + yk ds 1 = dx dy S1 F. ^nds = S1 (xi+yk).( k)dydx = 1 (ii) S (top): z = : ^n=k ds =dx dy S F.^ndS= S (x i+zj+yk).(k)dydx = 1 (iii) S 3 (right-hand end): y = 3; ^n=j ds 3 = dxdzj F = x i + zj + yk S3 F.^ndS= S3 (x i+zj+3k).(j)dzdx = 1 (iv) S 4 (left-hand end): y =, ^n = j, S4 F.^ndS = for S4 F.^ndS= S4 (x i+zj+yk).( j)dzdx= 1 = 1 ( ) dx = [xz]3 dx = 1 1x ] 1 dx=[6x = 6 3 ( y) dydx = 1 3 y dy dx = 9 y z dz dx = 1 ds 4 = dx dz y ( z) dz dx= 1 3 dx = 1 z dx = 9 dx = Now for the remaining two sides S 5 and S 6. Evaluate these in the same manner, obtaining S5 F.^n ds = 6; S5 F.^n ds = Check: dx 194

195 38-vector field (v) S 5 (front): x = 1; ^n = i ds 5 = dy dz S5 F. ^nds = S5 (i+zj+yk).(i)dy dz= S5 1 dy dz = 6 (vi) S 6 (back): x = ; ^n = i ds 6 =dy dz S6 F. ^nds = S6 (zj+yk).( i)dy dz = S6 dy dz = For the whole surface S we therefore have S F.dS = = 6 and from our previous work in section (a) V div F dv = 6 We have therefore verified as required that, in this example V div F dv = S F.dS Stokes Theorem If F is a vector field existing over an open surface S and around its boundary closed curve C, then S curl F.dS = O F.dr C This means that we can express a surface integral in terms of a line integral round the boundary curve. The proof of this theorem is rather lengthy and is to be found in the Appendix. Let us demonstrate its application in the following examples. Example A hemisphere S is defined by x + y + z = 4 (z > ). A vector field F = yi xj+xzk exists over the surface and around its boundary c. Verify Stoke s theorem that S curl F.dS = O C F.dr. S : x + y + z 4 = F = yi xj + xzk c is the circle x + y = 4. (a) O C F.dr= C (yi xj + xzk). (idx + jdy + kdz) = C (ydx x dy+xz dz) Converting to polar coordintes.x = cos θ; y = sin θ; z = dx = sin θ dθ; dy = cos θ dθ; Limits θ = to O F.dr = / (4sinθ[( sinθdθ] cosθcosθdθ = 4 / ( C sin θ + cos θ) dθ = 4 / (1 + sin θ) dθ = / (3 cos θ) dθ= sin θ 3θ = 1 (1) (b) Now we determine S curl F.dS curl F.dS = curl F. ^nds F = yi xj + xzk 195

196 38-vector field i j k curl F = x y z = i( ) j (z )+k ( 1 ) = zj 3k y x xz Now ^n= S S = xi+yj+zk 4x +4y +4z = xi+yj+zk Then S curl F. ^nds = S ( zk 3k). xi + yj + zk ds = 1 S ( yz 3z) ds x = sinφ cosθ; y = sinφ sinθ; z = cos φ, ds = 4 sin φ dφ dθ S curlf.^nds= 1 S ( sinφ sinθ cosφ 6 cosφ) 4 sinφ dφ dθ = / (sin φ cos φ sin θ +3sinφcosφ)dφdθ= 4 sin3 φ sinθ sin φ 3 sinθ + 3 dθ = 1 () So we have from our two results (1) and () S curl F.dS = O F.dr C = 4 / dθ 196

197 39-Periodic functions Lecture No -39 Periodic Functions Periodic functions A function f(x) is said to be periodic if its function values repeat at regular intervals of the independent variable. The regular interval between repetitions is the period of the oscillations. f(x + p) = f(x) Graphs of y = A sin nx (a) y = sin x The obvious example of a periodic function is y = sin x, which goes through its complete range of values while x increases from to 36. The period is therefore 36 or radians and the amplitude, the maximum displacement from the position of rest. y = 5sin x The amplitude is 5. The period is 18 and there are thus complete cycles in 36. Example Functions Amplitude Period y = 3 sin 5x 3 7 y = cos 3x 1 y = sin x 1 7 y = 4 sin x 4 18 y = A sin nx Thanking along the same lines, the function y = A sin nx has amplitude = A; period = 36 n = n ; n cycles in 36. Graphs of y = A cos nx have the same characteristics Example period = 8 ms period = 6 ms period = 5 cm Analytical description of a periodic function A periodic function can be defined analytically in many cases. Example (a) Between x = and x = 4, y = 3, i.e. f(x) = 3 < x < 4 (b) Between x = 4 and x = 6, y =, i.e. f(x) = 4 < x < 6. So we could define the function by f(x) = 3 < x < 4 197

198 39-Periodic functions f (x) = 4 < x < 6 f (x) = f (x + 6) the last line indicating that the function is periodic with period 6 units f(x) = x < x < 3 f(x) = 1 3 < x < 5 f(x) = f(x+5) f(x) = 3x 4 < x < 4 f(x) = 7 x 4 < x < 1 f(x) = 3 1 < x < 13 f(x) = f (x + 13) Example Sketch the graphs of the following inserting relevant values. 1. f(x) = 4 < x < 5 f(x) = 5 < x < 8 f(x) = f (x + 8) f(x) = 3x x < x < 3 f(x) = f (x + 3) f(x) = sin x < x < f(x) = < x < f(x) = f (x + ). f(x) = x 4 < x < 4 f(x) = 4 4 < x < 6 f(x) = 6 < x < 8 f(x) = f(x + 8). Useful integrals The following integrals appear frequently in our work on Fourier series, so it will help if we obtain the result in readiness. In each case, m and n are integers other than zero. (a) (b) sin nx dx = cos nx n cos nx dx = sin nx n = 1 n { cos n + cos n}= = 1 n {sinn + sin n} = (c) sin nx dx = 1 (1 cosnx) dx = 1 sin nx x n = (n ) 198

199 39-Periodic functions (d) (e) (f) cos nx dx = 1 (1 + cos nx) dx = 1 sin nx x + n = (n ) sin nx cos mx dx = 1 {sin(n+m)x+sin(n m)x}dx= 1 { + }= from result (a) with n m cos nx cos mx dx = 1 {cos(n+m)x+cos(n m)x}dx= 1 { + }= from result (b) with n m Note: If n = m, then (g) cos nx cos mx dx becomes sin nx sin mx dx = ½ cos nx dx = ( ) sin nx sin mxdx= ½ = 1 { )= from result (b) with n m from (d) above. cos(n+m)x cos(n m)xdx Note: If n = m, then sin nx sin mx dx becomes sin nx dx = from (c) above Summary of integrals (a) (d) (f) sin nx dx =, (b) cos nx dx = (n ) (e) cos nx cos mx dx = (n m) cos nx dx =, (c) sin nx cos mx dx = (g) cos nx cos mx dx = (n = m) sin nx dx = ( ) sin nx sin mx dx = (n m) sin nx sin mx dx = (n = m) Note We have evaluated the integrals between and, but, provided integration is carried out over a complete periodic interval of, the results are the same. Thus, the limits could just as well be to, to, / to 3/, etc. We can therefore choose the limits to suit the particular problems. 199

200 39-Periodic functions Fourier series Periodic functions of period The basis of a Fourier series is to represent a periodic function by a trigonometrical series of the form. f(x) = A +c 1 sin (x+α 1 ) + c sin (x+α ) +c 3 sin (3x+α 3 )+ +c n sin (nx+α n )+ where A is a constant term c 1, c, c 3.c n denote the amplitudes of the compound sine terms α 1, α, α 3.. are constant auxiliary angles. Each sine term, c n sin (nx + α n ) can be expanded thus: c n sin (nx + α n ) = c n {sin nx cosα n +cos n x sinα n }= (c n sinα n ) cos nx + (c n cosα n ) sin nx = a n cos nx + b n sin nx the whole series becomes. f(x) = A + {a n cos nx + b n sin nx} n=1

201 4-Fourier series Lecture No- 4 Fourier Series As we know that A + {a n cosnx + b n sinnx}; n=1 which can be written as in the expanded from A +(a 1 cosx+b 1 sinx)+(a cosx+b sinx) (a n cos nx+b n sin nx)+... A +a 1 cosx+a cosx+.+a n cos nx b 1 sinx+b sinx+... b n sin nx+... f(x) = A +a 1 cosx+a cosx+.+a n cos nx b 1 sinx+b sinx+... b n sin nx+... Fourier coefficients We have defined Fourier series in the form f(x) =A + n=1 {a n cosnx + b n sinnx}; n a positive integer (a) To find A, we integrate f(x) with respect to x from to. f(x)dx = A dx + n=1 a n cos nx dx+ b n sin nx dx = [A x] A = A = 1 f(x) dx f(x) dx =1/ a ; Where a = 1 f(x) dx (b) To find a n we multiply f(x) by cos mx and integrate from to. f(x)cos mx dx = f(x)cos mx dx = A {} + = + a n + = a n for n = m a n = 1 f(x) cos nx dx A cos mx dx+ n=1 a n cos nx cos mx dx+ b n sin nx cos mx dx {a n () + b n ()}= for n m n=1 (c) To find b n we multiply f(x) by sinmx and integrate from to. f(x)sin mx dx = f(x)sin mx dx = A {} + = + + b n = b n for n = m b n = 1 f(x) sin nx dx A sin mx dx + n=1 a n cosnxsinmx dx+ b n sin nx sinmxdx {a n () + b n ()}= for n m n=1 + { + } = A n=1 1

202 4-Fourier series Result For Fourier Series f(x)= 1 a + n=1 (a) a = 1 (b) a n = 1 (c) b n = 1 {a n cosnx+b n sinnx}; f(x) dx = mean value of f(x) over a period f(x) cos nx dx= mean value of f(x) cosnx over a period. f(x) sin nx dx = mean value of f(x) sin nx over a period. In each case, n = 1,, 3, Example Determine the Fourier series to represent the periodic function shown. It is more convenient here to take the limits as to. The function can be defined as f(x) = x < x < f(x) = f(x + ) period =. Now to find the coefficients (a) a = 1 f(x) dx= 1 x dx = 1 4 [x ] = a = (b) a n = 1 f(x) cos nx dx = 1 x cos nx dx a n = 1 x cos nx dx = 1 x sin nx [ n a n = (a) b n = 1 f(x) sin nx dx So we now have b n = 1 x 1 sin nx dx= [ x cosnx n ] ] a = ; a n = ; b n = 1 n Now the general expression for a Fourier series is f(x) = 1 a + 1 n sin nx dx = 1 ( ) 1 n () + 1 n 1 cos nx dx = n [ ]= 1 n {a n cos nx + b n sin nx} Therefore in this case n=1

203 4-Fourier series f(x) = + {b n sin nx} = +{ 1 1 sinx 1 sinx 1 3 sin3x...} since a n = n=1 f(x) = {sinx+1 sinx+1 3 sin3x+...} Dirichlet Conditions If the Fourier series is to represent a function f(x), then putting x = x 1 will give an infinite series in x 1 and the value of this should converge to the value of f(x 1 ) as more and more terms of the series are evaluated. For this to happen, the following conditions must be fulfilled. (a) The function f(x) must be defined and single-valued. (b) f(x) must be continuous or have a finite number of finite discontinuities within a periodic interval. (c) f(x) and f (x) must be piecewise continuous in the periodic interval. If these Dirichlet conditions are satisfied, the Fourier series converges to f(x 1 ), if x = x 1 is a point of continuity Example Find the Fourier series for the function shown. Consider one cycle between x= and x=. The function can be defined by f(x) = < x < f(x) = 4 < x < f(x) = < x < f(x) = f (x + ) f(x) = 1 a + {a n cos nx+b n sin nx} n=1 The expression for a is a = 1 f(x) dx This gives a = 1 / a = 4 (b) a n = 1 = 4 n / sin nx = dx + / / f(x) cos nx dx = 1 / 8 n sin n 4dx + dx = 1 / [4x] =4 / / / ()cosnxdx+ / / 4cosnxdx+ a n = 8 n sin n Then considering different integer values of n, we have If n is even a n = If n = 1.5, 9. a n = 8 n / () cosnxdx = 1 / / 4cosnxdx 3

204 4-Fourier series If n = 3, 7, 11,. a n = 8 n (c) To find b n b n = 1 f(x) sin nx dx = 1 / b n = 4 sin nx dx = 4 / cos nx n b n = / ()sinnxdx+ / / = 4 / / n 4sinnxdx+ () sinnxdx / cos n cos n = So with a = 4: an as stated above; b n = ; The Fourier series is f(x) = + 8 {cosx 1 3 cos 3x +1 5 cos5x 1 7 cos 7x +..} In this particular example, there are, in fact, no sine terms. Effect Of Harmonics It is interesting to see just how accurately the Fourier series represents the function with which it is associated. The complete representation requires an infinite number of terms, but we can, at least, see the effect of including the first few terms of the series. Let us consider the waveform shown. We established earlier that the function f(x) = < x < f(x) = 4 < x < f(x) = < x < f(x) = f (x + ) gives the Fourier series f(x) = + 8 {cos x 1 3 cos3x cos 5x 1 7 cos 7x +.} If we start with just one cosine term, we can then see the effect o including subsequent harmonics. Let us restrict our attention to just the right-hand half of the symmetrical waveform. Detailed plotting of points gives the following development. (1) f(x) = + 8 cos x 4

205 4-Fourier series () f(x) = + 8 {cosx 1 3 cos 3x} (3) f(x)=+ 8 {cos x 1 3 cos3x cos5x) (4) f(x) = + 8 {cosx 1 3 cos3x+ 1 3 cos5x 1 7 cos 7x} As the number of terms is increased, the graph gradually approaches the shape of the original square waveform. The ripples increase in number and decrease in amplitude, but a perfectly square waveform is unattainable in practice. For practical purpose, the first few terms normally suffice to give an accuracy of acceptable level. 5

206 41-Examples Lecture No -41 Examples Example Find the Fourier series for the function defined by f(x) = x < x < f(x) = < x < f(x) = f (x + ) The general expressions for a, a n, b n are a = 1 f(x) dx, a n = 1 f(x) cos nx dx, a = 1 f(x) dx = 1 (b) To find a n a n = 1 f(x) cos nx dx = 1 ( x) dx + 1 dx = 1 b n = 1 ( x) cosnxdx + 1 f(x) sin nx dx ( x) dx = 1 dx = 1 x = x cos nx dx = 1 sin nx [x n ] 1 n sin nx dx = 1 ( ) 1 n cos nx n = 1 1 n cos nx n = 1 cos nx n n = 1 n [cos cos n] = 1 n {1 cos n} But cos n = 1 (n even) and cos n = 1 (n odd) a n = n (n odd) and a n = (n even) (c) Now to find b n b n = 1 f(x)sinnxdx = 1 ( x) sin nx dx = 1 = 1 [x cos nx n ] x sin nx dx + 1 n cos nx dx = 1 cosn n + 1 n sin nx n b n = 1 n (n even) and b n = 1 n (n odd) So we have a = ; a n = (n even) and a n = n (n odd) cos n = n b n = 1 n (n even) and b n = 1 (n odd) n f(x)= 4 cosx+ 1 9 cos3x+ 1 5 cos 5x sin x 1 sin x sin 3x 1 4 sin 4x+. It is just a case of substituting n = 1,, 3, etc. In this particular example, we have a constant term and both sine and cosine terms. 6

207 41-Examples Odd And Even Functions (a) Even functions A function f(x) is said to be even if f( x) = f(x) i.e. the function value for a particular negative value of x is the same as that for the corresponding positive value of x. The graph of an even function is therefore symmetrical about the y-axis. y = f(x) = x is an even function since f( ) = 4 = f() f ( 3) = 9 = f(3) etc. y = f(x) = cos x is an even function since cos ( x) = cos x f( a) = cos a = f(a) (b) Odd functions A function f(x) is said to be odd if f ( x) = f(x) i.e. the function value for a particular negative value of xis numerically equal to that for the corresponding positive value of x but opposite in sign. The graph of an odd function is thus symmetrical about the origin. y = f(x) = x 3 is an odd function since f( ) = 8 = f() f ( 5) = 15 = f(5) etc. y = f(x) = sin x is an odd function Since sin ( x) = sin x f ( a) = f(a). So, for an even function f ( x) = f(x) symmetrical about the y-axis for an odd function f( x) = f (x) symmetrical about the origin. odd. Odd 7

208 41-Examples Even neither Even Odd Products Of Odd And Even Functions The rules closely resemble the elementary rules of signs. (even) (even)=(even) like (+) (+)=(+) ; (odd) (odd) = (even) ( ) ( )=(+) ; (odd) (even) = (odd) ( ) (+)=( ) The results can easily be proved. (a) Two even functions Let F(x) = f(x) g(x) where f(x) and g(x) are even functions. Then F( x) = f( x) g ( x) = f(x) g(x) since f(x) and g(x) are even. F( x) = F(x) F(x) is even (b) Two odd functions Let F(x) = f(x) g(x) where f(x) and g(x) are odd functions. Then F( x) = f( x) g( x) = { f (x)} { g(x)} since f(x) and g(x) are odd. = f(x) g(x) = F (x) F( x) = F(x) F (x) is even Finally (c) One odd and one even function Let F(x) = f(x) g(x) where f(x) is odd and g(x) even. Then F ( x) = f ( x) g( x) = f(x) g(x) = F (x) F ( x) = F (x) F (x) is odd So if f(x) and g(x) are both even, then f(x) g(x) is even and if f(x) and g(x) are both odd, then f(x) g(x) is even but if either f(x) or g(x) is even and the other odd. Then f(x) g(x) is odd. 8

209 41-Examples Example State whether each of the following products is odd, even, or neither. 1. x sin x odd (E) (O) = (O). x 3 cos x odd (O) (E) = (O) 3. cos x cos 3x even (E) (E) = (E) 4. x sin nx even (O) (O) = (E) 5. 3 sin x cos 4x odd (O) (E) = (O) 6. (x + 3) sin 4x neither (N) (O) = (N) 7. sin x cos 3 x even (E) (E) = (E) 8. x 3 e x neither (O) (N) = (N) 9. (x 4 + 4) sin x odd (E) (O) = (O) Two useful facts emerge from odd and even functions. (a) For an even function a f(x)dx= a a f(x) dx (b) For an odd function a a f (x) dx = THEOREM 1 If f(x) is defined over the interval < x < and f(x) is even, then the Fourier series for f(x) contains cosine terms only. Included in this is a which maybe regarded as a n cos nx with n =. Proof: (a) a = 1 f(x)dx = f(x)dx a = f(x) dx (b) a n = 1 f(x) cos nx dx. But f(x) cos nx is the product of two even functions and therefore itself even. a n = 1 f(x) cos nx dx = f(x) cos nx dx. a n = f(x) cos nx dx (c) b n = 1 f(x) sin nx dx Since f(x) sin nx is the product of an even function and an odd function, it is itself odd. b n = 1 f(x) sin nx dx =. b n = Therefore, there are no sine terms in the Fourier series for f(x). 9

210 41-Examples Example The waveform shown is symmetrical about the y-axis. The function is therefore even and there will be no sine terms in the series. f(x) = 1 a + (a) a = 1 n=1 a n cos nx f(x) dx = f(x) dx.= = / [4x] = 4 / 4 dx (b) a n = 1 f(x)cos nx dx = f(x) cos nx dx= 4cosnx dx= 8 sin nx n = 8 n sin n But sin n = for n even = 1 for n 1, 5, 9,.. = 1 for n = 3, 7, 11,. a n = (n even); a n = 8 n (n = 1, 5, 9,.); a n = 8 n (n = 3, 7, 11 ) (c) b n =, since f(x) is an even function.therefore, the required series is f(x) = + 8 {cos x 1 3 cos 3x cos 5x 1 7 cos 7x +.} Theorem : If f(x) is an odd function defined over the interval < x <, then the Fourier series for f(x) contains sine terms only. Proof: Since f(x) is an odd function f(x) dx = f(x) dx. (a) a = 1 f(x) dx. But f(x) is odd a = (b) a n = 1 f(x) cos nx dx Remembering that f(x) is odd and cosnx is even, the product f(x) cosnx is odd. a n = 1 f(x) cosnx dx = 1 a n = (odd function) dx / / 1

211 41-Examples (b) b n = 1 f(x) sin nx dx and since f(x) and sin nx are each odd, the product f(x) sin nx is even. Then b n = 1 (even function) dx = f(x) sin nx dx b n = f(x) sin nx dx So, If f(x) is odd function then a = ; a n = ; b n = f(x) sin nx dx i.e. the Fourier series contains sine terms only. Example f(x) = 6 < x < f(x) = 6 < x < f(x) = f(x + ) We can see that this is an odd function; a = and a n = b n = 1 f(x) sin nx dx. f(x) sin nx is a product of two odd functions and is therefore even. b n = f(x) sin nx dx b n = 6 sin nx dx = 1 cos nx n = 1 cos nx n = 1 [cos cos n] = 1 (1 cos n). n b n = (n even) b n = 4 (n odd) n So the series is f(x) = 4 {sinx sin3x sin5x +.} Of course, if f(x) is neither an odd nor an even function, then we must obtain expressions for a, a n and b n in full. 11

212 4-Examples Lecture No -4 Examples Example Determine the Fourier series for the function shown. This is neither odd nor even. Therefore we must find a, a n and b n. f(x) = 1 a + {a n cos nx + b n sin nx} n = 1 f(x) = x < x < =, < x < (a) a = 1 f(x)dx = 1 x dx + dx = 1 x [x] + = 1 { + 4 } a = 3 (b) a n = 1 f(x) cos nx dx = 1 x cos nx dx + cos nx dx = 1 x cosnx dx + cosnxdx = 1 x cosnx dx + cosnxdx = 1 x sin nx n 1 sinnxdx + cosnxdx n = 1 n cos nx n + sin nx n = 1 n cos nx n = n {cos n cos }= n {cos n 1} a n = (n even); a n = 4 n (n odd) (c) To find b n, we proceed in the same general manner b n = 1 f(x) sin nx dx= 1 x sin nx dx + sin nx dx 1 x cos nx = n cosnxdx+ sinnxdx + 1 n = 1 = 1 n cosn+( ) 1 n (cos n cosn) = = 1 n cos n = cos n n But cos n = 1. b n = n So the required series is n ( cosn)+ 1 n sin nx n + n cos nx 1 n cosn+( ) 1 n cos n+1 n cosn f(x) = 3 4 cosx+ 1 9 cos3x+ 1 5 cos5x+ sinx+ 1 sinx+1 3 sin3x+1 4 sin4x Sum of a Fourier series at a point of discontinuity 1

213 4-Examples f(x) = 1 a + {a n cos nx + b n sin nx} n=1 At x = x 1, the series converges to the value f(x 1 ) as the number of terms including increases to infinity. A particular point of interest occurs at a point of finite discontinuity or `jump of the function y = f(x). At x = x 1, the function appears to have two distinct values, y 1 and y. If we approach x = x 1 from below that value, the limiting value of f(x) is y 1. If we approach x = x 1 from above that value, the limiting value of f(x) is y. To distinguish between these two values we write y 1 = f(x 1 ) denoting immediately before x = x 1 y = f(x 1 + ) denoting immediately after x = x 1. In fact, if we substitute x = x 1 in the Fourier series for f(x), it can be shown that the series converges to the value 1 {f(x 1 ) + f(x 1 + )} = 1 (y 1 + y ), the average of y 1 and y. Example Consider the function f(x) = < x < f(x) = a f(x) = f (x + ) (a) a = 1 < x < f(x) dx= 1 a dx = 1 [ax] = a a = a (b) a n = 1 f(x)cos nx dx = 1 a cosnxdx = a sin nx n a n = b n = 1 f(x)sin nxdx = 1 a sin nx dx= a cos nx n = = a n (1 cos n) But cos n = 1 (n even) and cos n = -1 (n odd) b n = (n even) and b n = a n (n odd) 13

214 4-Examples f (x) = 1 a + n=1 b n sin nx f(x) = a + a {sinx sin3x sin5x +...} A finite discontinuity, or `jump, occurs at x =. If we substitute x = in the series obtained, all the sine terms vanish and we get f(x) = a/, which is, in fact, the average of the two function values at x =. Note also that at x =, another finite discontinuity occurs and substituting x = in the series gives the same result. Because of this ambiguity, the function is said to be `undefined at x =, x =, etc. Half-Range Series Sometime a function of period is defined over the range to, instead of the normal to, or to. We then have a choice of how to proceed. For example, if we are told between x = and x =, f(x) = x, then, since the period is, we have no evidence of how the function behave between x = and x =. If the waveform were as shown in (a), the function would be an even function, symmetrical about the y-axis and the series would have only cosine terms (including possibly a ). On the other hand, if the waveform were as shown in (b), the function would be odd, being symmetrical about the origin and the series would have only sine terms. Example A function f(x) is defined by f(x) = x < x < f(x) = f(x + ) Obtain a half-range cosine series to represents the function. To obtain a cosine series, i.e. a series with no sine terms, we need an even function. Therefore, we assume the waveform between x = and x = to be as shown, making the total graph symmetrical about the y-axis. Now we can find expressions for the Fourier coefficients as usual. a = a = f(x)dx = xdx = [x ] = 14

215 4-Examples a n = 1 f(x) cos nx dx = f(x) cos nx dx a n = = 4 x cos nx dx ( ) 1 n cos nx n = 4 x cos nx dx = 4 x sin nx n = ( 4 n ) (cos n 1) 1 n sin nx dx cos n = 1 (n even) and cos n = 1 (n odd) a n = (n even) and a n = 8 n (n odd) All that now remains is b n which is zero, since f(x) is an even function, i.e. b n = So a =, a n = (n even) and a n = 8 n (n odd), b n =. Therefore f(x)= 8 cosx+ 1 9 cos3x+ 1 5 cos5x+. Example Determine a half-range sine series to repressent the function f(x) defined by f(x) = 1 + x < x < f(x) = f(x + ) We choose the waveform between x = and x = so that the graph is symmetrical about the origin. The function is then an odd function and the series will contain only sine terms. a = and a n = b n = 1 f(x) sin nx dx b n = f(x) sin nx dx b n = (1+x)sin nx dx = (1 + x) cosnx n + 1 n cosnxdx = 1 + n cos n + 1 n +1 n sin nx n = 1 n 1 + n cos n = {1 (1 + ) cos n} n cos n = 1 (n even) and cos n = 1 (n odd) b n = n = 4+ n (n even) (n odd) Substituting in the general expression f(x) = b n sin nx we have x = 1 15

216 4-Examples f(x)= 4 + {sinx sin3x+1 5 sin5x+...} {1 sin x+1 4 sin 4x+1 6 sin 6x+...) and the required series obtained f(x)= 4 + {sin x+1 3 sin 3x+1 5 sin 5x+..} {1 sin x+1 4 sin 4x+1 6 sin 6x+...} So knowledge of odd and even functions and of half-range series saves a deal of unnecessary work on occasions. 16

217 43-Functions with periods other than Lecture No.-43 Functions With Periods Other Than So far, we have considered functions f(x) with period. In practice, we often encounter functions defined over periodic intervals other than, e.g. from to T, T to T etc. Functions With Period T If y = f(x) is defined in the range T to T, i.e. has a period T, we can convert this to an interval of by changing the units of the independent variable. In many practical cases involving physical oscillations, the independent variable is time (t) and the periodic interval is normally denoted by T, i.e. f(t) = f(t + T) Each cycle is therefore completed in T seconds and the frequency f hertz (oscillations per second) of the periodic function is therefore given by f = 1 T. If the angular velocity, ω radians per seconds, is defined by ω = f, then ω = T and T = ω The angle, x radians, at any time t is therefore x = ωt and the Fourier series to represent the function can be expressed as f(t) = 1 a + {a n Cos nωt +b n Sin nωt} x=1 which can also be written in the form f(t) = 1 A + x=1 B n sin (nωt + φ n ) n = 1,, 3,... Fourier Coefficients With the new variable, the Fourier coefficients become: f(t) = 1 a + x=1 a = T T a n = T T {a n cosnωt + b n sinnωt} f(t) dt = ω /ω f(t) dt f(t) cos nωt dt = ω /ω f(t)cos nωt dt b n = T T f(t) sin nωt dt= ω /ω f(t)sin nωt dt We can see that there is very little difference between these expressions and those that have gone before. The limits can, of course, be to T, T to T, ω to ω, to ω etc. as is convenient, so long as they cover a complete period. Example Determine the Fourier series for a periodic function defined by 17

218 43-Functions with periods other than f(t) = (1 + t) 1 < t < f(t) = < t < 1 f(t) = f (t + ) The first step is to sketch the wave which is. f(t) = 1 a + x=1 T = a = T T/ T/ {a n cos nωt+b n sin nωt}) f(t) dt = 1 f(t) dt = (1 + t) dt + 1 () dt= [t + t ] = a = 1 a n = T/ T f(t)cosnωtdt = T/ a n = (1+t) sin nωt nω = n ω (1 cos nω) Now T = ω nω 1 ω = T = = a n = Now for b n a n = (n even) 4 = n ω (n odd) b n = T T/ T/ f(t) sin nωt dt f(t)cosnωt dt= 1 (1 + t) cosnωt dt sinnωtdt = ( ) 1 nω cos nωt nω n ω (1 cos n) b n = (1+t) sin nωt dt = 1 (1+t) cos nωt nω nω 1 = 1 nω + 1 nω sin nωt nω = 1 1 nω + 1 n ω (sin nω) As before ω = b n = nω So the first few terms of the series give cosnωtdt f(t)= ω cos ωt+ 1 9 cos3ωt+ 1 5 cos5ωt+.. ω sin ωt+ 1 sinωt+1 3 sin3ωt+1 4 sin4ωt

219 43-Functions with periods other than Half-Range Series The theory behind the half-range sine and cosine series still applies with the new variable. (a) Even function Half-range cosine series y = f(t) < t < T f(t) = f(t + T) symmetrical about the y-axis. With an even function, we know that b n = f(t) = 1 a + a n cos nωt n=1 where a = 4 T/ T f(t) dt and (b) Odd function Half-range sine series y = f(t) < t < T f(t) = f(t + T) symmetrical about the origin. a = and a n = f(t) = b n sin nωt; x=1 b n = 4 T/ T f(t) sin nωt dt a n = 4 T T/ f(t) cos nωt dt Example A function f(t) is defined by f(t) = 4 t, < t < 4. We have to form a half-range cosine series to represent the function in this interval. First we form an even function, i.e. symmetrical about the y-axis. a = 4 T/ T f 1 (t) dt = (4 t) dt= 1 4 (4 t) dt = 1 4t t a n = 4 T/ T 4 = 1 4 (4) (4) = 1 [16 8]= 1 (8)= 4 f 1 (t)cos nωt dt = (4 t) cos nωt dt Simple integration by parts gives a n = 1 sin 4nω 1 nω n ω (cos 4nω 1) But ω = T = 8 = 4 a n = 1 sinn nω 1 n ω (cos n 1) n = 1,, 3,.. sin n = ; 19

220 43-Functions with periods other than cos n = 1 (n even); cos n = 1 (n odd) a n = (n even) 1 a n = n ω (n odd) f(t)=+ 1 ω cos ωt+ 1 9 cos3ωt+ 1 5 cos5ωt+ where ω = 4. Example A function f(t) is defined by f(t) = 3 + t < t < f(t) = f(t + 4) Obtain the half-range sine series for the function in this range. Sine series required. Therefore, we form an function, symmetrical about the origin. a = ; a n = ; T = 4 f(t) = b n sin nωt n=1 b n = 4 T/ T f(t) sin nωt dt= (3 + t) sin nωt dt (3 + t) sin nωt dt = (3+t)cosnωt n ω cos nωt nω dt (3 + ) cos nω 3 = nω nω + 1 nω sin nωt n ω = 3 nω 5 nω cos nω+ 1 n ω sin nω nω But T = ω = ω T = = 3 nω 5 nω cos n+ 1 n ω sin n nω b n = 1 nω (3 5 cosnω) + 1 n ω (sin nω) b n = 1 nω (3 5cosn)+ 1 n ω (sin n) = nω = 8 nω (n even) (n odd) odd f(t)= ω 4sinωt 1 sinωt+4 3 sin3ωt 1 4 sin4ωt.. Half-Wave Rectifier A sinusoidal voltage E sin ωt, where t is time, is passed through a half-wave rectifier that clips the negative portion of the wave Find the Fourier series of the resulting periodic functions. u(t) = if T/ < t < = E sin ωt < t < T/ here T = ω